ESAT Chemistry Mock Module 2 Worked Solutions

Correct Answer: C. Using the same mass of marble in larger chips

Fastest Approach (🚀):
Larger chips $\Rightarrow$ less surface area $\Rightarrow$ slower.

Matches Option C.

Step-by-Step Breakdown:

1. The mechanism behind every rate factor

A reaction proceeds when particles collide with enough energy. Anything that increases how often they collide, or how energetically, increases the rate.

2. Test each change

Powder: far greater total surface area, so many more acid particles are in contact with solid at once — faster.
Higher concentration: more acid particles per unit volume, so more collisions per second — faster.
Higher temperature: particles move faster, colliding both more often and with more energy — faster, and this is the largest effect of the four.
Larger chips: less total surface area for the same mass, so fewer collisions per second — slower.
Catalyst: provides a route with lower activation energy, so a greater proportion of collisions succeed — faster.

3. Surface area without changing the amount

The mass of marble is the same in Option D; only its division changes. One large lump has far less exposed surface than the same mass ground fine, and only the surface can react. Everything inside a chip is unreachable until the outside has dissolved away.

4. Where this matters outside the laboratory

Flour, custard powder and coal dust are all combustible as powders and safe as solids, for exactly this reason. Dust explosions in mills are surface area taken to its conclusion.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Grinding the marble into a powder — Does Increase
  Powder has a much larger surface area — faster.
• B. Increasing the concentration of the acid — Does Increase
  More particles per unit volume — faster.
• D. Raising the temperature — Does Increase
  More frequent and more energetic collisions — faster.
• E. Adding a suitable catalyst — Does Increase
  Lowers the activation energy — faster.

Common Mistake (⚠️):
Assuming the same mass means the same rate. Only the exposed surface reacts, so how finely the solid is divided changes the rate without changing the amount.

Takeaway (📌):
Concentration, temperature, surface area and catalysts all raise the rate. Larger pieces cut the surface area and slow it down.

Correct Answer: E. Because the electrodes must keep a fixed polarity, so that each ion is always attracted to the same electrode and its product collects there

Fastest Approach (🚀):
AC reverses the electrodes many times a second.
Products would form and re-dissolve at both electrodes.

Matches Option E.

Step-by-Step Breakdown:

1. The terms

Electrode — a conducting rod dipping into the electrolyte.
Cathode — the negative electrode, which attracts the positive ions (cations).
Anode — the positive electrode, which attracts the negative ions (anions).
Electrolyte — a molten or dissolved ionic compound in which the ions are free to move.

2. Why the polarity must stay fixed

Cations are attracted to the cathode and are discharged there; anions go to the anode. With direct current those roles never change, so the two products form at separate electrodes and can be collected separately.

Alternating current reverses the polarity fifty times a second. Each electrode would alternately attract cations and anions, so any product formed would immediately be reversed at the same electrode. Nothing would accumulate anywhere and no separation would occur.

3. Why the other options fail

Alternating current is not weaker, and it flows through an electrolyte perfectly well — so Options A and D are simply wrong. And the electrolyte's ions already exist; electrolysis does not create them, it moves and discharges them, which disposes of Option E.

4. The memory aid

CATions to the CAThode. The cathode is negative in electrolysis, which is worth fixing because it is positive in a discharging battery, and the two contexts are easily confused.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Because alternating current is too weak to decompose the electrolyte — Factually Wrong
  Alternating current is not inherently weaker.
• B. Because alternating current cannot ionise the electrolyte — Mechanism Wrong
  The ions exist already in a molten or dissolved ionic compound.
• C. Because alternating current would heat the electrolyte too much — Wrong Cause
  Heating is not the reason electrolysis needs direct current.
• D. Because only direct current can flow through a liquid — Factually Wrong
  Alternating current flows through electrolytes readily.

Common Mistake (⚠️):
Assuming the cathode is positive because the name sounds like it should be. In electrolysis the cathode is the negative electrode, and cations go to it.

Takeaway (📌):
Cathode negative, anode positive, cations to the cathode. Direct current keeps those roles fixed so the products separate.

Correct Answer: C. A chemical process such as electrolysis, because the elements are chemically bonded

Fastest Approach (🚀):
Compound, not mixture $\Rightarrow$ chemical process needed.
Aluminium is above carbon in the reactivity series $\Rightarrow$ electrolysis.

Matches Option C.

Step-by-Step Breakdown:

1. Compound or mixture

Aluminium oxide is a compound: aluminium and oxygen are held by strong ionic bonds in a fixed ratio. It is not a mixture of the two elements.

2. Why every physical technique fails

Filtration, distillation, evaporation and chromatography all exploit differences in physical properties between substances that are merely mixed. None of them can break a chemical bond, so none can release aluminium from its oxide.

3. Which chemical process

Aluminium sits above carbon in the reactivity series, so it cannot be displaced by heating with carbon as iron can. It is extracted by electrolysis of the molten oxide, dissolved in cryolite to lower the operating temperature.

4. Why aluminium was once precious

Electrolysis needs a great deal of electrical energy, so until cheap electricity existed aluminium could not be extracted at scale. It was more valuable than gold in the mid-nineteenth century — a fact that follows directly from its position above carbon in the reactivity series.

5. The general rule

Above carbon: electrolysis.
Below carbon: reduction by heating with carbon.
Very unreactive, such as gold: found uncombined, needing no extraction at all.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Filtration, because the oxide is insoluble — Physical Method
  Filtration separates an insoluble solid from a liquid, not elements from a compound.
• B. Fractional distillation, because aluminium and oxygen have different boiling points — Physical Method
  Distillation separates mixtures of liquids.
• D. Evaporation, because the oxygen escapes as a gas on heating — Physical Method
  Evaporation removes a solvent; it does not decompose the oxide.
• E. Chromatography, because the two elements travel at different rates — Physical Method
  Chromatography separates mixtures of solutes.

Common Mistake (⚠️):
Treating a compound as though it were a mixture. Physical separation techniques never break chemical bonds, however the components are described.

Takeaway (📌):
Mixtures separate physically; compounds need a chemical reaction. Metals above carbon need electrolysis, those below need carbon reduction.

Correct Answer: E. Calcium chloride, water and carbon dioxide

Fastest Approach (🚀):
Carbonate $\Rightarrow$ salt, water and carbon dioxide.

Matches Option E.

Step-by-Step Breakdown:

1. Identify the reaction type

Calcium carbonate is a carbonate, so this is the acid-plus-carbonate reaction.

2. The three general equations

acid $+$ metal $\rightarrow$ salt $+$ hydrogen
acid $+$ base or alkali $\rightarrow$ salt $+$ water
acid $+$ carbonate $\rightarrow$ salt $+$ water $+$ carbon dioxide

3. Apply it

$2\text{HCl}+\text{CaCO}_3\rightarrow\text{CaCl}_2+\text{H}_2\text{O}+\text{CO}_2$

4. Naming the salt

The metal comes from the carbonate and the rest of the name from the acid: hydrochloric gives chlorides, sulfuric gives sulfates, nitric gives nitrates. So calcium carbonate with hydrochloric acid gives calcium chloride.

5. The observation

Effervescence — the fizzing is carbon dioxide escaping. Bubbling that gas through limewater turns it milky, which is the standard test and confirms the third product. Option A omits precisely the product you can see.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Calcium chloride and water only — Product Omitted
  Correct for a base, but a carbonate also gives carbon dioxide.
• B. Calcium chloride and oxygen — Wrong Product
  Oxygen is not produced.
• C. Calcium chloride and hydrogen — Wrong Reaction Type
  Hydrogen is produced with a metal, not a carbonate.
• D. Calcium hydroxide and carbon dioxide — Wrong Product
  Calcium hydroxide is not a product of this reaction.

Common Mistake (⚠️):
Giving only salt and water, as for a base. The carbonate reaction has a third product, and it is the visible one.

Takeaway (📌):
Metal gives hydrogen, base gives water, carbonate gives water and carbon dioxide. The salt's name comes from the metal plus the acid.

Correct Answer: B. Metal layers can slide while the delocalised electrons keep holding the ions together; displaced ionic layers bring like charges together, which repel

Fastest Approach (🚀):
Metals: delocalised electrons hold sliding layers together.
Ionic: displaced layers put like charges adjacent $\Rightarrow$ repulsion $\Rightarrow$ shatter.

Matches Option B.

Step-by-Step Breakdown:

1. The metallic structure

A giant lattice of positive metal ions in a sea of delocalised electrons — outer electrons no longer belonging to any particular atom, free to move through the whole structure.

2. Why metals are malleable

The attraction between the positive ions and the electron sea is not directional. Push one layer of ions past another and the electrons simply flow with them, so the attraction is undiminished. The metal changes shape without breaking.

3. Why ionic solids shatter

An ionic lattice is a strict alternation of positive and negative. Displace one layer by a single ion's width and positive ions come to face positive ions. The electrostatic repulsion splits the crystal along that plane, so the solid shatters rather than bending.

4. Conduction, from the same fact

The delocalised electrons are also free to carry charge, which is why metals conduct electricity as solids. Ionic compounds do not — their ions are locked in place — but they conduct once molten or dissolved, because then the ions can move.

One structural feature, three properties: malleability, conduction, and the contrast with ionic behaviour.

5. Why Option A fails

Metallic bonding is strong, not weak — that is why metals have high melting points. The malleability comes from the bonding being non-directional, not from it being feeble.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Metallic bonds are weaker, so the layers separate easily — Strength Misstated
  Metallic bonds are strong; high melting points show it.
• C. Metals are held by weak intermolecular forces that reform after distortion — Structure Wrong
  Metals are giant structures, not molecular ones.
• D. Ionic solids contain no bonds between layers — Factually Wrong
  Ionic attraction acts throughout the lattice, including between layers.
• E. Metals contain no ions, so there is nothing to repel — Factually Wrong
  Metals do contain positive ions.

Common Mistake (⚠️):
Explaining malleability by saying metallic bonds are weak. They are strong; what matters is that they act equally in all directions and so survive the layers moving.

Takeaway (📌):
Metals: positive ions in a delocalised electron sea, non-directional, so malleable and conducting. Ionic: alternating charges, so displacement causes repulsion and shattering.

Correct Answer: E. They have full outer electron shells, so they have no tendency to lose, gain or share electrons

Fastest Approach (🚀):
Full outer shell $\Rightarrow$ nothing to gain by reacting.

Matches Option E.

Step-by-Step Breakdown:

1. The configurations

Helium: $2$ — a full first shell.
Neon: $2,8$.
Argon: $2,8,8$.

Each has a complete outer shell.

2. Why that means unreactive

Atoms react to reach a stable, full outer shell. The noble gases already have one, so losing, gaining or sharing electrons would move them away from the stable arrangement rather than towards it. There is no driving force.

3. Why this is the reference point for everything else

Every other element's chemistry is described by how it reaches a noble gas configuration. Sodium loses one electron to become like neon; chlorine gains one to become like argon. The noble gases define the target, which is why they sit at the end of each Period.

4. Why the other options fail

They are monatomic, not diatomic, so Option C is wrong on the facts. They certainly have outer electrons — full shells of them — so Option D is wrong. And plenty of gases react vigorously, hydrogen and oxygen among them, so Option E proves nothing.

5. The uses that follow

Argon fills light bulbs and shields welding precisely because it will not react with the hot metal. Helium fills balloons because it is light and will not burn — unlike hydrogen, which is lighter still.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. They are very heavy, so their atoms move too slowly to react — Wrong Cause
  Helium is very light and equally unreactive.
• B. They are all gases, and gases do not react — Wrong Cause
  Hydrogen and oxygen are gases and react readily.
• C. They exist as diatomic molecules, which are very stable — Factually Wrong
  Noble gases are monatomic.
• D. They have no electrons in their outer shell — Reversed
  They have full outer shells, not empty ones.

Common Mistake (⚠️):
Saying noble gases have no outer electrons. They have full outer shells, which is the opposite of having none, and it is the fullness that makes them stable.

Takeaway (📌):
Full outer shell means no tendency to react. Every other element's chemistry is an attempt to reach that same configuration.

Correct Answer: D. Petrol, because its short molecules have weak intermolecular forces and a low boiling point

Fastest Approach (🚀):
Top is coolest $\Rightarrow$ lowest boiling point $\Rightarrow$ shortest chains.

Matches Option D.

Step-by-Step Breakdown:

1. How the column works

Crude oil is heated and vaporised at the base. The column is hot at the bottom and progressively cooler towards the top. Each vapour rises until it reaches a level cool enough for it to condense, and is drawn off there.

2. What sets the boiling point

Hydrocarbon molecules are held to one another by intermolecular forces. Longer chains have more contact between molecules and so stronger forces, needing more energy to separate.

Short chains — weak forces, low boiling point, condense high in the column. Refinery gases and petrol.
Long chains — strong forces, high boiling point, condense low. Diesel, fuel oil and bitumen.

3. Apply it

Petrol has among the shortest molecules of the named fractions, so it condenses near the top.

4. The properties that follow chain length

Going down the column the fractions become more viscous, less volatile, darker, and harder to ignite. All of those follow from the same increase in intermolecular forces, which is why one idea explains the whole column.

5. Why cracking exists

Demand for short-chain fractions such as petrol exceeds their natural abundance in crude oil, while long-chain fractions are in surplus. Cracking breaks long molecules into shorter ones, matching supply to demand and producing alkenes as a valuable by-product.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Bitumen, because it has the longest molecules and the highest boiling point — Direction Reversed
  A high boiling point means condensing low, not high.
• B. Bitumen, because light fractions sink — Wrong Mechanism
  Fractions separate by boiling point, not by sinking.
• C. Diesel, because it has an intermediate boiling point — Position Wrong
  Diesel condenses in the middle-to-lower region.
• E. Petrol, because its molecules are chemically the most stable — Wrong Cause
  Chemical stability is not what determines boiling point.

Common Mistake (⚠️):
Confusing the covalent bonds inside a molecule with the forces between molecules. Boiling separates molecules, so it is the intermolecular forces that set the boiling point.

Takeaway (📌):
Short chains, weak forces, low boiling point, top of the column. Every property of a fraction follows from its chain length.

Correct Answer: B. A reactant has been used up, so the reaction has stopped

Fastest Approach (🚀):
Flat $\Rightarrow$ gradient zero $\Rightarrow$ rate zero $\Rightarrow$ reaction over.

Matches Option B.

Step-by-Step Breakdown:

1. Gradient is rate

On a graph of product against time, the gradient at any point is the rate of reaction at that moment.

2. Read the three regions

Steepest at the start — concentrations are highest, so collisions are most frequent and the rate is greatest.

Gradually shallower — reactants are being consumed, so concentrations fall and collisions become less frequent. The rate decreases continuously.

Level — the gradient is zero, so no more product is forming. A reactant has been completely used up and the reaction has finished.

3. What the final height tells you

The total volume of gas at the plateau is fixed by the limiting reactant. Repeating with more of that reactant raises the plateau; repeating with a catalyst or at a higher temperature reaches the same plateau sooner — the reaction goes faster but produces no more product.

That distinction, height against steepness, is what these graphs are usually testing.

4. Why Option E misreads it

A constant rate would be a straight line with a non-zero gradient. A horizontal line means the rate is zero, not steady.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. The reaction has slowed because the temperature has fallen — Wrong Cause
  Most such reactions are exothermic and warm slightly; concentration is what falls.
• C. The gas has begun to dissolve in the solution — Wrong Cause
  Would reduce the measured volume, not level it off.
• D. The reaction has reached equilibrium and is now reversible — Unsupported
  Nothing in the description indicates a reversible reaction.
• E. The rate is constant and the reaction continues at a steady speed — Gradient Misread
  A horizontal line means zero rate, not a constant one.

Common Mistake (⚠️):
Reading a flat line as a steady rate. Flat means the gradient is zero, so the rate is zero and the reaction has stopped.

Takeaway (📌):
Gradient is rate. Steep start, shallowing as reactants are consumed, flat when the limiting reactant runs out. The plateau height depends only on the amount of that reactant.

Correct Answer: D. cathode $\text{Pb}^{2+}+2\text{e}^{-}\rightarrow\text{Pb}$; anode $2\text{Br}^{-}\rightarrow\text{Br}_2+2\text{e}^{-}$

Fastest Approach (🚀):
Cathode: reduction, electrons in. Anode: oxidation, electrons out.
$\text{Pb}^{2+}$ to the cathode, $\text{Br}^{-}$ to the anode.

Matches Option D.

Step-by-Step Breakdown:

1. Send each ion to the right electrode

$\text{Pb}^{2+}$ is positive, so it travels to the negative cathode.
$\text{Br}^{-}$ is negative, so it travels to the positive anode.

That alone eliminates Option C, which has them the wrong way round.

2. Write the cathode half-equation

The cathode supplies electrons, so the cation gains them — reduction:
$\text{Pb}^{2+}+2\text{e}^{-}\rightarrow\text{Pb}$

Two electrons, because the charge must balance: $+2$ and $-2$ give zero.

3. Write the anode half-equation

The anion loses electrons — oxidation. Bromine atoms then pair into a molecule, so two bromide ions are needed:
$2\text{Br}^{-}\rightarrow\text{Br}_2+2\text{e}^{-}$

4. Check the electron counts match

Two electrons released at the anode, two consumed at the cathode. They must be equal, since the same current passes through both — a useful check on any pair of half-equations.

5. Why the other options fail

Option B has the electrons on the wrong side at both electrodes. Option D uses one electron for a $2+$ ion, so the charge does not balance, and produces atomic rather than molecular bromine. Option E oxidises lead metal that is not present as a reactant.

6. What is observed

A silvery bead of molten lead collects at the cathode, and orange-brown bromine vapour is released at the anode.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. cathode $\text{Pb}^{2+}+\text{e}^{-}\rightarrow\text{Pb}$; anode $\text{Br}^{-}\rightarrow\text{Br}+\text{e}^{-}$ — Charge Unbalanced
  Charges do not balance, and bromine is diatomic.
• B. cathode $\text{Pb}^{2+}\rightarrow\text{Pb}+2\text{e}^{-}$; anode $\text{Br}_2+2\text{e}^{-}\rightarrow 2\text{Br}^{-}$ — Electrons Misplaced
  Electrons on the wrong side at both electrodes.
• C. cathode $2\text{Br}^{-}\rightarrow\text{Br}_2+2\text{e}^{-}$; anode $\text{Pb}^{2+}+2\text{e}^{-}\rightarrow\text{Pb}$ — Electrodes Swapped
  The two electrodes swapped.
• E. cathode $\text{Pb}\rightarrow\text{Pb}^{2+}+2\text{e}^{-}$; anode $2\text{Br}^{-}\rightarrow\text{Br}_2+2\text{e}^{-}$ — Wrong Species
  Oxidises lead metal, which is a product rather than a reactant.

Common Mistake (⚠️):
Putting the electrons on the wrong side. At the cathode electrons are gained, so they appear on the left; at the anode they are lost and appear on the right.

Takeaway (📌):
Cathode: reduction, electrons on the left. Anode: oxidation, electrons on the right. Balance the charges and pair up diatomic products.

Correct Answer: C. A separating funnel

Fastest Approach (🚀):
Immiscible $\Rightarrow$ two layers $\Rightarrow$ separating funnel.

Matches Option C.

Step-by-Step Breakdown:

1. Establish miscibility

Oil and water do not mix. Left to stand they form two distinct layers, with the less dense oil floating on the water.

2. Exploit the layers

A separating funnel holds the mixture with a tap at the bottom. Open the tap and the lower layer — the water — runs out first; close it as the boundary reaches the tap, and the oil is left behind.

Simple, and it needs no heating at all.

3. Why the other techniques do not fit

Fractional distillation is for miscible liquids with different boiling points, such as ethanol and water or the fractions of crude oil. Using it here would work but is needless: it requires heating a mixture that separates on standing.

Filtration separates an insoluble solid from a liquid. Both components here are liquids, so nothing is retained by the paper.

Chromatography separates dissolved solutes by how strongly they adsorb.

Crystallisation recovers a dissolved solid from its solution.

4. The decision rule

Solid in liquid, insoluble: filter.
Solid dissolved: evaporate or crystallise.
Liquids, immiscible: separating funnel.
Liquids, miscible: fractional distillation.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Fractional distillation — Wrong Technique
  For miscible liquids with different boiling points.
• B. Filtration — Wrong Technique
  Separates an insoluble solid from a liquid.
• D. Chromatography — Wrong Technique
  Separates dissolved solutes.
• E. Crystallisation — Wrong Technique
  Recovers a dissolved solid from solution.

Common Mistake (⚠️):
Reaching for distillation for any pair of liquids. It is the right answer only when they are miscible; immiscible liquids separate themselves and need only draining.

Takeaway (📌):
Immiscible liquids: separating funnel. Miscible liquids: fractional distillation. The miscibility decides it before anything else.

Correct Answer: A. An alkali is a soluble base, so all alkalis are bases but not all bases are alkalis

Fastest Approach (🚀):
Alkali $=$ soluble base.
All alkalis are bases; not all bases are alkalis.

Matches Option A.

Step-by-Step Breakdown:

1. The definitions

A base is a substance that neutralises an acid — an $\text{H}^{+}$ acceptor. Metal oxides, metal hydroxides and metal carbonates are bases.

An alkali is a base that is soluble in water, and therefore able to release $\text{OH}^{-}$ ions into solution.

2. The direction of the relationship

Every alkali is a base, because dissolving does not stop it neutralising acid. But many bases are insoluble and so are not alkalis.

Copper(II) oxide is a base — it neutralises acid to give copper(II) sulfate — but it is insoluble, so it is not an alkali. Sodium hydroxide is both.

3. Why Option C is wrong

Donating $\text{H}^{+}$ is the definition of an acid. A base accepts it. Reversing the two is the standard error, and it is what Option C is built on.

4. Why the distinction is practical

Only an alkali gives an alkaline solution, so only an alkali has a pH above 7 in water and only an alkali turns an indicator blue. An insoluble base still neutralises acid perfectly well — it simply cannot do so in solution until the acid dissolves it.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. All bases are alkalis — Direction Reversed
  Reverses the subset — many bases are insoluble.
• C. A base donates hydrogen ions in solution — Definition Reversed
  Donating hydrogen ions defines an acid.
• D. Bases and alkalis are two names for the same thing — Distinction Denied
  They differ in solubility.
• E. An alkali is an insoluble base — Reversed
  An alkali is a soluble base, not an insoluble one.

Common Mistake (⚠️):
Treating base and alkali as interchangeable. The difference is solubility, and it determines whether the substance can act in solution at all.

Takeaway (📌):
Base: neutralises acid, accepts $\text{H}^{+}$. Alkali: a base that dissolves in water. Alkalis are a subset of bases.

Correct Answer: B. $\text{CO}_2$ is simple molecular, so only weak intermolecular forces need to be overcome; $\text{SiO}_2$ is a giant covalent structure, so strong covalent bonds must be broken

Fastest Approach (🚀):
$\text{CO}_2$: simple molecular, weak forces between molecules.
$\text{SiO}_2$: giant covalent, bonds must break.

Matches Option B.

Step-by-Step Breakdown:

1. Identify each structure

Carbon dioxide — discrete $\text{CO}_2$ molecules. Strong double covalent bonds within each molecule, but only weak intermolecular forces between molecules.

Silicon dioxide — a giant covalent (macromolecular) structure, in which every silicon is covalently bonded to four oxygens throughout the whole crystal. There are no separate molecules at all.

2. Ask what melting actually breaks

Melting or boiling separates the particles from one another. For carbon dioxide, the particles are molecules, so only the weak forces between them must be overcome — the $\text{C}=\text{O}$ bonds inside each molecule survive intact into the gas phase.

For silicon dioxide, separating the particles means breaking covalent bonds, because those are what hold one part of the structure to the next.

3. Why this is the whole answer

Both substances have strong covalent bonds, so Option A is simply false. The difference is not bond strength but what has to be broken, which is determined by the structure.

4. The general rule

Simple molecular: low melting and boiling points, does not conduct.
Giant covalent: very high melting points, usually does not conduct — graphite excepted, because of its delocalised electrons.
Giant ionic: high melting point, conducts when molten or aqueous.
Metallic: high melting point, conducts when solid.

Naming the structure gives every physical property.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. The covalent bonds in $\text{CO}_2$ are much weaker than those in $\text{SiO}_2$ — Bond Strength
  Both have strong covalent bonds; the bonds are not the difference.
• C. $\text{CO}_2$ is ionic and $\text{SiO}_2$ is covalent — Bond Type Wrong
  $\text{CO}_2$ is covalent, not ionic.
• D. $\text{CO}_2$ molecules are lighter, and lighter substances always boil lower — Wrong Cause
  Mass alone does not determine boiling point; structure does.
• E. $\text{SiO}_2$ contains metallic bonding as well as covalent bonding — Factually Wrong
  Silicon dioxide contains no metallic bonding.

Common Mistake (⚠️):
Saying covalent bonds break when carbon dioxide boils. Gaseous $\text{CO}_2$ is still $\text{CO}_2$ — its bonds are intact, and only the forces between molecules were overcome.

Takeaway (📌):
Name the structure first. Simple molecular melts by overcoming weak intermolecular forces; giant structures melt by breaking bonds.

Correct Answer: A. Potassium's outer electron is in a shell further from the nucleus and is shielded by more inner shells, so it is lost more easily

Fastest Approach (🚀):
Further out and better shielded $\Rightarrow$ easier to lose.

Matches Option A.

Step-by-Step Breakdown:

1. What Group 1 reactivity depends on

All alkali metals react by losing their single outer electron. So the more easily that electron is lost, the more reactive the metal.

2. The two competing effects going down the Group

Nuclear charge rises — lithium has $3$ protons, potassium $19$. On its own this would hold the outer electron more tightly.

Distance and shielding rise — lithium's outer electron is in the second shell, potassium's in the fourth, with more inner shells between it and the nucleus repelling it outwards.

3. Which wins

Distance and shielding win. The outer electron of potassium experiences a weaker net attraction despite the larger nucleus, so it is lost more easily and potassium is more reactive.

4. Why the trend reverses for the halogens

Halogens react by gaining an electron, so the same structural change has the opposite effect: further out and better shielded means the incoming electron is attracted less strongly, so reactivity falls down Group 17. One structural argument, two opposite trends — which is exactly why the two Groups must not be conflated.

5. Why Option C is the sharp distractor

The nuclear charge does increase, and stated alone it predicts the wrong trend. A complete answer must say that shielding and distance outweigh it.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Potassium has more outer electrons to lose — Factually Wrong
  All Group 1 metals have exactly one outer electron.
• C. Potassium has a greater nuclear charge, which attracts the outer electron more strongly — Incomplete Reasoning
  True but predicts the opposite trend on its own.
• D. Potassium is denser, so its atoms collide more often — Wrong Cause
  Density is not what determines reactivity here.
• E. Potassium has fewer protons than lithium — Factually Wrong
  Potassium has more protons, not fewer.

Common Mistake (⚠️):
Citing the increased nuclear charge as the reason for greater reactivity. That effect alone predicts the opposite trend; it is outweighed by distance and shielding.

Takeaway (📌):
Down Group 1: bigger atoms, more shielding, outer electron lost more easily, reactivity rises. Down Group 17 the same changes make reactivity fall.

Correct Answer: E. $\text{C}_5\text{H}_{12}$

Fastest Approach (🚀):
$2(5)+2 = 12$, so $\text{C}_5\text{H}_{12}$.

Matches Option E.

Step-by-Step Breakdown:

1. Apply the general formula

Alkanes are $\text{C}_n\text{H}_{2n+2}$. With $n = 5$:
$2(5)+2 = 12$

so the formula is $\text{C}_5\text{H}_{12}$, pentane.

2. Why the plus two

Each carbon in the chain forms four bonds. Along the chain, carbons bond to their neighbours; the two carbons at the ends each have one spare bond, and those two extra hydrogens are the $+2$.

3. Why alkanes are called saturated

Every carbon-carbon bond is a single bond, so the molecule holds the maximum possible hydrogen. Nothing more can be added without removing something — the molecule is saturated with hydrogen.

4. The contrast with alkenes

Alkenes are $\text{C}_n\text{H}_{2n}$, two hydrogens fewer, because one $\text{C}=\text{C}$ double bond replaces two $\text{C}-\text{H}$ bonds. $\text{C}_5\text{H}_{10}$ is therefore pentene, and it appears here as Option A.

5. What a homologous series is

A family sharing a general formula, with each member differing from the next by $\text{CH}_2$. Members show a gradual trend in physical properties and similar chemical reactions — which is why learning one formula covers the whole series.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\text{C}_5\text{H}_{10}$ — Wrong Series
  The alkene formula — two hydrogens short.
• B. $\text{C}_5\text{H}_5$ — Formula Error
  Far too few hydrogens for a saturated chain.
• C. $\text{C}_5\text{H}_{11}$ — Group Not Molecule
  An alkyl group, not a complete molecule.
• D. $\text{C}_5\text{H}_8$ — Formula Error
  Four hydrogens short of an alkane.

Common Mistake (⚠️):
Using $\text{C}_n\text{H}_{2n}$ and giving $\text{C}_5\text{H}_{10}$. That is the alkene formula; alkanes carry two more hydrogens.

Takeaway (📌):
Alkanes $\text{C}_n\text{H}_{2n+2}$, alkenes $\text{C}_n\text{H}_{2n}$, alcohols $\text{C}_n\text{H}_{2n+1}\text{OH}$. Two hydrogens separate the first two.

Correct Answer: C. Because heating increases both the frequency of collisions and the proportion that exceed the activation energy, whereas concentration affects only the frequency

Fastest Approach (🚀):
Temperature: more collisions and more energetic ones.
Concentration: more collisions only.

Matches Option C.

Step-by-Step Breakdown:

1. What concentration does

More particles in the same volume means they meet more often. The collision frequency rises, but the energy of each collision is unchanged, so the proportion that succeed is the same.

2. What temperature does

Particles gain kinetic energy, so they move faster. That has two separate consequences:

They collide more often — same effect as concentration.
A greater proportion of collisions exceed the activation energy — this is the additional effect, and it is much the larger one.

3. Why the second effect dominates

Only collisions with at least the activation energy can react. At a typical temperature only a small fraction qualify, so a modest rise in temperature moves a disproportionately large number of particles past the threshold. A rise of $10\ ^{\circ}\text{C}$ roughly doubles the rate of many reactions — far more than a $10\%$ rise in concentration achieves.

4. What temperature does not do

It does not lower the activation energy — that is what a catalyst does, by providing an alternative pathway. Temperature leaves the barrier where it is and gives more particles the energy to clear it. Option C confuses the two mechanisms, which is the sharpest distractor here.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Because heating increases the number of particles present — Factually Wrong
  Heating does not create particles.
• B. Because heating lowers the activation energy of the reaction — Mechanism Confused
  A catalyst lowers activation energy; temperature does not.
• D. Because heating makes the particles larger, so they collide more easily — Factually Wrong
  Particles do not change size on heating.
• E. Because concentration has no effect on the rate of reaction — Effect Denied
  Concentration does affect the rate.

Common Mistake (⚠️):
Saying heating lowers the activation energy. It raises the energy of the particles; only a catalyst lowers the barrier.

Takeaway (📌):
Concentration raises collision frequency. Temperature raises frequency and the fraction above the activation energy, which is why it matters more.

Correct Answer: A. hydrogen at the cathode, chlorine at the anode

Fastest Approach (🚀):
Cathode: $\text{Na}$ more reactive than $\text{H} \Rightarrow$ hydrogen.
Anode: halide present $\Rightarrow$ chlorine.

Matches Option A.

Step-by-Step Breakdown:

1. List every ion present

From the salt: $\text{Na}^{+}$ and $\text{Cl}^{-}$.
From the water: $\text{H}^{+}$ and $\text{OH}^{-}$.

Both electrodes therefore have two candidates.

2. Decide at the cathode

$\text{Na}^{+}$ and $\text{H}^{+}$ both arrive. The less reactive metal is discharged, and hydrogen sits below sodium in the reactivity series. So hydrogen is produced:
$2\text{H}^{+}+2\text{e}^{-}\rightarrow\text{H}_2$

Sodium is far too reactive to be deposited from aqueous solution — it would immediately react with the water.

3. Decide at the anode

$\text{Cl}^{-}$ and $\text{OH}^{-}$ both arrive. When a halide is present in reasonable concentration, the halogen is discharged in preference to oxygen:
$2\text{Cl}^{-}\rightarrow\text{Cl}_2+2\text{e}^{-}$

With a sulfate or nitrate instead, no halide is available and oxygen is released from the hydroxide ions.

4. What is left behind

$\text{Na}^{+}$ and $\text{OH}^{-}$ remain in solution as sodium hydroxide. Electrolysing brine therefore yields three useful products at once — hydrogen, chlorine and sodium hydroxide — which is the basis of the chlor-alkali industry.

5. Contrast with the molten compound

Electrolyse molten sodium chloride and there is no water, so no competition: sodium metal forms at the cathode and chlorine at the anode. That is Option A, and it is the right answer to a different question.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. sodium at the cathode, chlorine at the anode — Molten Not Aqueous
  The products from **molten** sodium chloride, where water is absent.
• C. hydrogen at the cathode, oxygen at the anode — Halide Ignored
  Correct for a sulfate or nitrate, but a halide is present here.
• D. sodium at the cathode, oxygen at the anode — Water Ignored
  Both predictions ignore the competing water species.
• E. chlorine at the cathode, hydrogen at the anode — Electrodes Swapped
  Electrodes swapped — chlorine is negative and goes to the anode.

Common Mistake (⚠️):
Predicting the metal at the cathode from an aqueous solution of a reactive metal. In water, hydrogen is discharged instead unless the metal is below hydrogen in the reactivity series.

Takeaway (📌):
Aqueous: cathode gives hydrogen unless the metal is below it in the reactivity series; anode gives the halogen if a halide is present, otherwise oxygen. Molten: the elements of the compound.

Correct Answer: A. Simple distillation, condensing the evaporated water

Fastest Approach (🚀):
Want the water $\Rightarrow$ must condense the vapour $\Rightarrow$ distillation.

Matches Option A.

Step-by-Step Breakdown:

1. Identify which component is wanted

The question asks for pure water — the solvent. That single word decides the technique.

2. Why distillation

Heating the solution evaporates the water; the salt has a far higher boiling point and stays behind. The vapour passes into a condenser, cools, and is collected as pure liquid water in a separate vessel.

3. Why evaporation and crystallisation fail here

Both drive the water off into the air and keep the salt. They are the right answer to the opposite question — recover the dissolved solid — but here they discard exactly what was wanted. Option A also has it backwards: crystallisation leaves the salt behind, not the water.

4. Why filtration fails

Salt is dissolved, so its ions pass straight through the filter paper with the water. Filtration only holds back particles that are undissolved.

5. Simple against fractional

Simple distillation suffices when only one component is volatile, as here. Fractional distillation, with a column, is needed when two liquids both evaporate and their boiling points are close — ethanol and water, or the fractions of crude oil.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Filtration, which holds back the salt — Dissolved Not Suspended
  Dissolved salt passes through filter paper.
• C. Crystallisation, which leaves the water behind — Wrong Component
  Crystallisation keeps the salt and loses the water.
• D. Evaporation to dryness — Wrong Component
  Evaporation loses the water to the air.
• E. Chromatography — Wrong Technique
  Chromatography separates solutes, not solvent from solute.

Common Mistake (⚠️):
Choosing evaporation because it separates salt from water. It does — but it keeps the salt and loses the water, which is the reverse of what was asked.

Takeaway (📌):
Want the solvent: distil and condense. Want the solute: evaporate or crystallise. The wanted component picks the technique.

Correct Answer: D. $\text{H}^{+}(\text{aq})+\text{OH}^{-}(\text{aq})\rightarrow\text{H}_2\text{O}(\text{l})$

Fastest Approach (🚀):
Cancel the spectator ions.
Only $\text{H}^{+}$ and $\text{OH}^{-}$ react.

Matches Option D.

Step-by-Step Breakdown:

1. Write the full equation with ions

$\text{H}^{+}+\text{Cl}^{-}+\text{Na}^{+}+\text{OH}^{-}\rightarrow\text{Na}^{+}+\text{Cl}^{-}+\text{H}_2\text{O}$

2. Cancel the spectators

$\text{Na}^{+}$ and $\text{Cl}^{-}$ appear unchanged on both sides — they remain dissolved throughout, and take no part. Cancelling them leaves:

$\text{H}^{+}(\text{aq})+\text{OH}^{-}(\text{aq})\rightarrow\text{H}_2\text{O}(\text{l})$

3. Why this is the general answer

Neither the acid's anion nor the alkali's cation appears. Nitric acid with potassium hydroxide, or sulfuric acid with sodium hydroxide, reduces to exactly the same equation — which is why the enthalpy of neutralisation is essentially identical for all strong acid–strong alkali pairs, at about $-57\ \text{kJ mol}^{-1}$.

4. Why the state symbol matters

Water is formed as a liquid, not a gas — the reaction happens in solution at room temperature. Option D changes only that symbol, and it is wrong.

5. Why the reaction warms the mixture

Bond formation in water releases more energy than is needed to separate the ions, so the reaction is exothermic and the temperature rises — which is what makes neutralisation measurable in a calorimeter.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\text{HCl}(\text{aq})+\text{NaOH}(\text{aq})\rightarrow\text{NaCl}(\text{aq})+\text{H}_2\text{O}(\text{l})$ — Not Ionic
  The full molecular equation, with spectators still present.
• B. $\text{H}^{+}(\text{aq})+\text{OH}^{-}(\text{aq})\rightarrow\text{H}_2\text{O}(\text{g})$ — State Symbol
  Water forms as a liquid in solution, not a gas.
• C. $\text{Na}^{+}(\text{aq})+\text{Cl}^{-}(\text{aq})\rightarrow\text{NaCl}(\text{aq})$ — No Change
  Describes the salt remaining dissolved, which is no reaction at all.
• E. $2\text{H}^{+}(\text{aq})+\text{O}^{2-}(\text{aq})\rightarrow\text{H}_2\text{O}(\text{l})$ — Species Wrong
  Free $\text{O}^{2-}$ ions do not exist in aqueous solution.

Common Mistake (⚠️):
Quoting the full molecular equation when the ionic one was asked for. An ionic equation must have the spectator ions removed.

Takeaway (📌):
Strong acid plus strong alkali always reduces to $\text{H}^{+}+\text{OH}^{-}\rightarrow\text{H}_2\text{O}(\text{l})$, whatever the salt formed.

Correct Answer: B. Giant ionic

Fastest Approach (🚀):
Conducts molten but not solid $\Rightarrow$ ionic.

Matches Option B.

Step-by-Step Breakdown:

1. Use the melting point

High, so the particles are strongly held throughout the structure. That rules out simple molecular, which has only weak intermolecular forces and melts low.

2. Use the solid conductivity

It does not conduct as a solid, so there are no mobile charge carriers in the solid. That rules out metallic, where delocalised electrons conduct in the solid state, and it rules out graphite for the same reason.

3. Use the molten conductivity

It conducts when molten, so charge carriers exist and become mobile on melting. Those carriers must be ions: fixed in position in the lattice, free to move in the liquid.

That rules out ordinary giant covalent structures such as diamond and silicon dioxide, which have no charged particles at all and conduct in neither state.

4. The conclusion

Only a giant ionic structure fits all three observations. It would also dissolve in water to give a conducting solution, for the same reason.

5. The table this question is built from

| structure | melting point | solid | molten |
|---|---|---|---|
| simple molecular | low | no | no |
| giant covalent | very high | no | no |
| giant ionic | high | no | yes |
| metallic | high | yes | yes |

Every combination of the last two columns names exactly one structure, which is why conductivity is the most informative single test.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Simple molecular — Melting Point
  Simple molecular substances melt at low temperatures.
• C. Giant covalent — Conductivity
  Giant covalent structures conduct in neither state.
• D. Metallic — Solid Conductivity
  A metal conducts as a solid.
• E. Giant covalent with delocalised electrons, like graphite — Solid Conductivity
  Graphite conducts as a solid, because of its delocalised electrons.

Common Mistake (⚠️):
Choosing metallic because the substance conducts. It conducts only when molten; a metal conducts as a solid too, and that distinction is what the question turns on.

Takeaway (📌):
Solid no, molten yes means ionic. Both yes means metallic. Neither, with a high melting point, means giant covalent.

Correct Answer: E. $\text{Cl}_2+2\text{NaBr}\rightarrow 2\text{NaCl}+\text{Br}_2$

Fastest Approach (🚀):
Reactivity: $\text{Cl}_2>\text{Br}_2>\text{I}_2$.
Only chlorine displacing bromide runs downhill.

Matches Option E.

Step-by-Step Breakdown:

1. The reactivity order

Reactivity decreases down Group 17:
$\text{F}_2>\text{Cl}_2>\text{Br}_2>\text{I}_2$

Halogens react by gaining an electron, and the smaller atoms higher in the Group attract that electron more strongly.

2. The displacement rule

A halogen displaces the halide of any halogen below it. Chlorine can take the electron away from a bromide ion, but bromine cannot take it from a chloride ion.

3. Test the options

$\text{I}_2$ with chloride: iodine is below chlorine — no reaction.
$\text{Br}_2$ with chloride: bromine is below chlorine — no reaction.
$\text{Cl}_2$ with bromide: chlorine is above bromine — reaction occurs ✓
$\text{I}_2$ with bromide: iodine is below bromine — no reaction.

4. What is observed

The colourless solution turns orange as bromine is released. The ionic equation strips out the spectator sodium:
$\text{Cl}_2+2\text{Br}^{-}\rightarrow 2\text{Cl}^{-}+\text{Br}_2$

It is a redox reaction: chlorine is reduced and bromide oxidised.

5. Why the trend opposes Group 1

Group 1 metals become more reactive down the Group because losing an electron gets easier. Halogens become less reactive down the Group because gaining one gets harder. Same structural cause, opposite consequence.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\text{I}_2+2\text{NaCl}\rightarrow 2\text{NaI}+\text{Cl}_2$ — Wrong Direction
  Iodine is less reactive than chlorine.
• B. $\text{Br}_2+2\text{NaCl}\rightarrow 2\text{NaBr}+\text{Cl}_2$ — Wrong Direction
  Bromine is less reactive than chlorine.
• C. None of these, because halogens do not displace one another — Reaction Denied
  Displacement between halogens is a standard reaction.
• D. $\text{I}_2+2\text{NaBr}\rightarrow 2\text{NaI}+\text{Br}_2$ — Wrong Direction
  Iodine is less reactive than bromine.

Common Mistake (⚠️):
Applying the Group 1 trend to the halogens. Reactivity rises down Group 1 and falls down Group 17, and confusing them reverses every displacement prediction.

Takeaway (📌):
Halogen reactivity falls down the Group. A halogen displaces halides of those below it, never above.

Correct Answer: E. It provides an alternative pathway of lower activation energy, and is chemically unchanged at the end

Fastest Approach (🚀):
Lower activation energy, unchanged at the end.
Rate changes; yield does not.

Matches Option E.

Step-by-Step Breakdown:

1. The mechanism

A catalyst provides an alternative reaction pathway with a lower activation energy. More collisions therefore have enough energy to react, so the rate rises.

2. What is unchanged

The catalyst is not consumed. It may take part in intermediate steps, but it is regenerated, so it is chemically unchanged at the end and a small amount catalyses a large quantity of reaction indefinitely.

3. What else is unchanged

The energies of the reactants and products are untouched, so:
the enthalpy change $\Delta H$ is the same
the position of equilibrium is the same
the final yield is the same

Only the height of the barrier between them falls. Options A and D both claim the endpoint changes, and both are wrong for the same reason.

4. On the energy level diagram

The reactant and product levels stay exactly where they were. Only the peak comes down, and the difference between the two levels — which is $\Delta H$ — is unaffected.

5. Why they matter industrially

The Haber process uses iron; catalytic converters use platinum and rhodium. A catalyst lets a reaction run acceptably fast at a lower temperature, which saves energy and cost — the point being speed, never quantity.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. It increases the yield of product as well as the rate — Yield Misconception
  Yield is unchanged; only the rate rises.
• B. It is used up during the reaction and must be replaced — Consumption Misconception
  A catalyst is regenerated and not consumed.
• C. It raises the temperature of the reaction mixture — Mechanism Wrong
  A catalyst does not heat the mixture.
• D. It lowers the energy of the products, making the reaction more exothermic — Enthalpy Misconception
  Product energies are unchanged, so $\Delta H$ is unchanged.

Common Mistake (⚠️):
Believing a catalyst increases the yield. It changes only how fast equilibrium or completion is reached, never how much product forms.

Takeaway (📌):
A catalyst lowers the activation energy and is unchanged at the end. Rate rises; $\Delta H$, equilibrium position and yield do not.

Correct Answer: C. Spoon as the cathode, silver as the anode, in a solution containing silver ions

Fastest Approach (🚀):
Metal deposits at the cathode $\Rightarrow$ spoon is the cathode.
Silver anode dissolves to replenish the solution.

Matches Option C.

Step-by-Step Breakdown:

1. Which electrode the object must be

Silver ions carry a positive charge, so they migrate to the negative electrode — the cathode — where they gain electrons and deposit as silver metal:
$\text{Ag}^{+}+\text{e}^{-}\rightarrow\text{Ag}$

For the silver to land on the spoon, the spoon must be the cathode.

2. What the anode should be

Pure silver. As plating proceeds, the anode dissolves:
$\text{Ag}\rightarrow\text{Ag}^{+}+\text{e}^{-}$

replacing exactly the silver ions removed at the cathode. The electrolyte's concentration therefore stays constant and the process can run indefinitely.

3. What the electrolyte must contain

A soluble silver salt, such as silver nitrate, so that $\text{Ag}^{+}$ ions are available. A solution of iron ions would deposit iron, not silver, which is what Option C gets wrong.

4. Why plate at all

Appearance, and protection. A thin layer of an unreactive metal shields the cheaper, more reactive metal beneath from corrosion, at a fraction of the cost of making the whole object from silver.

5. Why Option A fails twice

With the spoon as the anode it would dissolve rather than gain a coating — the exact opposite of the intention.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Spoon as the anode, silver as the cathode, in a silver salt solution — Electrodes Swapped
  As the anode the spoon would dissolve, not be coated.
• B. Spoon as the cathode, silver as the anode, in a solution containing iron ions — Wrong Electrolyte
  Iron ions would deposit iron, not silver.
• D. Both electrodes made of silver, with the spoon suspended between them — Not An Electrode
  An object not connected as an electrode is not plated.
• E. Spoon as the anode, carbon as the cathode, in a silver salt solution — Electrodes Swapped
  Spoon as anode again, and a carbon cathode would collect the silver.

Common Mistake (⚠️):
Making the object the anode. The anode dissolves; the cathode receives the deposit, so the object being plated must always be the cathode.

Takeaway (📌):
Object as cathode, plating metal as anode, electrolyte containing that metal's ions. The anode dissolves to keep the solution constant.

Correct Answer: A. The dye is a mixture of at least three substances

Fastest Approach (🚀):
Three spots $\Rightarrow$ at least three substances $\Rightarrow$ a mixture.

Matches Option A.

Step-by-Step Breakdown:

1. How chromatography separates

The solvent rises through the paper carrying the dissolved substances with it. Each substance is carried a different distance depending on how strongly it is attracted to the paper against how soluble it is in the solvent. Separate substances therefore end at separate heights.

2. Reading the result

One spot: a single substance — pure.
More than one spot: more than one substance — a mixture.

Three spots means the dye contains at least three different substances.

3. Why 'at least'

Two substances with very similar attractions can travel together and appear as one spot. So the spot count is a minimum, not necessarily the exact number of components. Repeating with a different solvent can resolve them.

4. Why 'pure' and 'compound' are not the same

Option C conflates them. A compound is a single substance and gives one spot however many elements it contains — water is one compound of two elements and shows a single spot. Purity in chromatography means one substance, not one element.

5. The R f value

Dividing the distance moved by the substance by the distance moved by the solvent front gives the $R_f$ value, which is characteristic of a substance in a given solvent and lets each spot be identified rather than merely counted.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. The dye is a pure substance with three components — Purity Confused
  A pure substance gives one spot.
• C. The dye is a compound of three elements — Compound vs Mixture
  A compound is one substance and gives one spot.
• D. The experiment failed, because a pure dye should give three spots — Result Misread
  Three spots is a valid result, not a failure.
• E. The dye is pure, and the extra spots are the solvent — Factually Wrong
  The solvent front is not a spot; it is the line the solvent reaches.

Common Mistake (⚠️):
Confusing a pure compound with a mixture. A compound of several elements is still one substance and gives one spot; three spots means three substances.

Takeaway (📌):
One spot means pure. Several spots means a mixture, and the count is a lower bound because spots can overlap.

Correct Answer: C. It is exothermic, with $\Delta H$ negative, and the surroundings warm

Fastest Approach (🚀):
Products lower $\Rightarrow$ energy released $\Rightarrow$ exothermic, $\Delta H<0$, surroundings warm.

Matches Option C.

Step-by-Step Breakdown:

1. Read the diagram

The products sit below the reactants, so the system has lost energy. That energy has gone to the surroundings.

2. The three consequences, in order

Energy released $\rightarrow$ the reaction is exothermic
$\Delta H$ measures the change in the system, which has fallen $\rightarrow$ $\Delta H$ is negative
The energy has gone to the surroundings $\rightarrow$ they warm up

3. Why the sign feels counter-intuitive

The mixture in the beaker gets hotter, which feels like a gain — but $\Delta H$ refers to the chemicals, not the thermometer. The system lost the energy that the surroundings gained, so the sign is negative even though the temperature rises. That inversion is what Options C and E exploit.

4. The endothermic mirror

Products above reactants: energy absorbed from the surroundings, $\Delta H$ positive, and the surroundings cool. Dissolving ammonium nitrate in water is the standard demonstration.

5. What else the diagram shows

The peak between reactants and products is the activation energy, and it exists for both types. An exothermic reaction can still need a large input to get started — which is why methane is safe until it is lit.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. It is endothermic, with $\Delta H$ positive, and the surroundings cool — Direction Reversed
  Describes the opposite case entirely.
• B. It is exothermic, with $\Delta H$ positive, and the surroundings warm — Sign Wrong
  Right type, but exothermic means a negative ΔH.
• D. It has no enthalpy change, because energy is conserved — System vs Universe
  Energy is conserved overall, but it has moved out of the system.
• E. It is endothermic, with $\Delta H$ negative, and the surroundings warm — Internally Inconsistent
  Internally inconsistent: endothermic means a positive ΔH.

Common Mistake (⚠️):
Making $\Delta H$ positive because the surroundings get hotter. $\Delta H$ describes the system, and the system has lost the energy the surroundings gained.

Takeaway (📌):
Products lower: exothermic, $\Delta H$ negative, surroundings warm. Products higher: endothermic, $\Delta H$ positive, surroundings cool.

Correct Answer: A. $8.4\ \text{kJ}$

Fastest Approach (🚀):
$100\times 4.2 = 420$, then $\times 20 = 8400\ \text{J} = 8.4\ \text{kJ}$.

Matches Option A.

Step-by-Step Breakdown:

1. Apply the equation

$q = mc\Delta T = 100\times 4.2\times 20 = 8400\ \text{J}$

2. Convert

$8400\ \text{J} = 8.4\ \text{kJ}$

Both forms appear among the options, so read which unit is wanted.

3. Which mass to use

The water's mass, not the fuel's. The calorimetry equation measures the energy absorbed by the water, and the water's specific heat capacity is what is quoted. Substituting the mass of fuel burnt is the standard error.

4. From this to the enthalpy of combustion

Divide by the moles of fuel burnt to get the energy per mole, and attach a negative sign because combustion is exothermic. If $0.01\ \text{mol}$ burnt here, then
$\Delta H = -\dfrac{8.4}{0.01} = -840\ \text{kJ mol}^{-1}$

5. Why experimental values come out low

Heat is lost to the surroundings and to the apparatus, and combustion may be incomplete. A school calorimetry result is typically well below the accepted value, and always in the same direction — which is a systematic error rather than a random one.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $840\ \text{J}$ — Decimal Slip
  A factor of ten lost.
• C. $84\ \text{kJ}$ — Conversion Error
  A factor of ten gained in the conversion.
• D. $2100\ \text{J}$ — Substitution Error
  Using $\Delta T = 5$, or omitting a factor.
• E. $0.84\ \text{kJ}$ — Conversion Error
  Two factors of ten lost.

Common Mistake (⚠️):
Using the mass of fuel instead of the mass of water. The equation describes what the water absorbed, which is what the temperature rise measures.

Takeaway (📌):
$q = mc\Delta T$, with $m$ the mass of water. Divide by moles of fuel and negate for the enthalpy of combustion.

Correct Answer: D. $-184\ \text{kJ mol}^{-1}$

Fastest Approach (🚀):
Broken $= 678$; formed $= 862$.
$678-862 = -184\ \text{kJ mol}^{-1}$.

Matches Option D.

Step-by-Step Breakdown:

1. Bonds broken, in the reactants

One $\text{H}-\text{H}$ and one $\text{Cl}-\text{Cl}$:
$436+242 = 678\ \text{kJ mol}^{-1}$ absorbed.

2. Bonds formed, in the products

The equation makes two $\text{HCl}$ molecules, each with one $\text{H}-\text{Cl}$ bond:
$2\times 431 = 862\ \text{kJ mol}^{-1}$ released.

3. Subtract

$\Delta H = \text{broken}-\text{formed} = 678-862 = -184\ \text{kJ mol}^{-1}$

4. Check the sign against the chemistry

More energy was released forming bonds than was absorbed breaking them, so the reaction is exothermic and $\Delta H$ is negative. Hydrogen burning in chlorine does indeed release energy, so the sign is right.

5. The two errors this question is built around

Forgetting the coefficient gives $678-431 = +247$, which appears as Option D with the sign error and Option C without. And reversing the subtraction to formed minus broken gives $+184$ — Option A — which would make an obviously exothermic reaction endothermic.

6. Why the result is approximate

Tabulated bond energies are averages over many compounds, so a calculated $\Delta H$ is close to but not exactly the measured value.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $+184\ \text{kJ mol}^{-1}$ — Order Reversed
  Formed minus broken — the subtraction reversed.
• B. $+247\ \text{kJ mol}^{-1}$ — Two Errors
  Coefficient dropped and the subtraction reversed.
• C. $-247\ \text{kJ mol}^{-1}$ — Coefficient Dropped
  Only one $\text{H}-\text{Cl}$ bond counted.
• E. $-862\ \text{kJ mol}^{-1}$ — Incomplete
  The bonds formed alone, not the difference.

Common Mistake (⚠️):
Computing formed minus broken. Breaking costs and forming pays, so it is broken minus formed — and the sign check against whether the reaction warms up catches the reversal.

Takeaway (📌):
$\Delta H = \sum\text{bonds broken}-\sum\text{bonds formed}$. Multiply each bond energy by the number of those bonds in the balanced equation.

Correct Answer: D. $+78\ \text{kJ mol}^{-1}$

Fastest Approach (🚀):
Forward releases $\Rightarrow -78$.
Reverse $\Rightarrow +78\ \text{kJ mol}^{-1}$.

Matches Option D.

Step-by-Step Breakdown:

1. Sign the forward reaction

Energy is released, so hydration is exothermic:
$\Delta H = -78\ \text{kJ mol}^{-1}$

2. Reverse it

Going back to the anhydrous salt must absorb exactly what was released, so the reverse reaction is endothermic with the same magnitude:
$\Delta H = +78\ \text{kJ mol}^{-1}$

3. Why the magnitude is identical

The two states — anhydrous plus water, and hydrated — sit at fixed energy levels. $\Delta H$ is the difference between them, and reversing the direction of travel changes only which one you start from. The gap itself is unchanged.

4. What is actually observed

Adding water to white anhydrous copper(II) sulfate turns it blue and the test tube becomes noticeably warm — the exothermic direction, and the standard test for water.

Heating the blue hydrated salt drives the water off and returns it to white, and that requires continuous heating because it is endothermic. Both facts come from the same enthalpy value with opposite signs.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $-78\ \text{kJ mol}^{-1}$ — Sign Not Flipped
  The forward reaction's value, not the reverse.
• B. $+156\ \text{kJ mol}^{-1}$ — Magnitude Wrong
  The magnitude is unchanged, not doubled.
• C. $0\ \text{kJ mol}^{-1}$ — Misconception
  The two directions differ; they do not cancel within one reaction.
• E. $-39\ \text{kJ mol}^{-1}$ — Magnitude Wrong
  The magnitude is unchanged, not halved.

Common Mistake (⚠️):
Keeping the same sign for the reverse reaction. The magnitude is conserved but the direction of energy flow inverts, so the sign must flip.

Takeaway (📌):
Reversing a reaction reverses the sign of $\Delta H$ and keeps its magnitude. Exothermic one way means endothermic the other.