ESAT Practice Set 1A Chemistry Worked Solutions

Correct Answer: A. 12

Fastest Approach (🚀):
Compute $mc = 200 \times 4.2 = 840\ \text{J}$ per degree first and write the heat straight in joules as $10080$. Faster still, the mass in grams is numerically equal to the volume in $\text{cm}^3$, so the volume cancels and the rise is just $0.2 \times 252 \div 4.2 = 12$, with no litres and no grams to carry.

Step-by-Step Breakdown:

1. Find the amount of manganese displaced

The magnesium is in excess, so every manganese ion reacts and the manganese(II) sulfate is the limiting reagent. The volume in cubic decimetres is $0.2\ \text{dm}^3$, so

The reaction is $\mathrm{Mg} + \mathrm{Mn^{2+}} \rightarrow \mathrm{Mg^{2+}} + \mathrm{Mn}$, one manganese atom for each manganese ion, so $0.04\ \text{mol}$ of manganese is displaced.

2. Convert that amount into a total heat

3. Turn the heat into a temperature rise

What warms up is the $200\ \text{cm}^3$ of solution, and at a density of $1\ \text{g cm}^{-3}$ that is $200\ \text{g}$. The metal's mass is ignored and the cup loses no heat, so all $10080\ \text{J}$ goes into the solution. Rearranging $Q = mc\Delta T$,

Sanity check: $840\ \text{J}$ raises this solution by one degree, and $10080$ is $12$ times $840$, so $12$ degrees. Notice that the volume never reaches the answer: it fixes the amount reacting and the mass warmed in the same proportion, so it cancels, and the rise depends only on the concentration, the energy per mole and the specific heat capacity.

The temperature rise is $12\ ^\circ\text{C}$.

Why the Other Options Are Wrong (❌):

• B. 1.2 · Volume conversion off by ten
  The volume was converted the wrong way, $200\ \text{cm}^3$ taken as $0.02\ \text{dm}^3$, giving $n = 0.004\ \text{mol}$ and $Q = 1.008\ \text{kJ}$, then $1008 \div 840 = 1.2$. A $\text{dm}^3$ is a thousand $\text{cm}^3$, so the volume is $0.2\ \text{dm}^3$ and the amount reacting is ten times larger than this.
• C. 6 · Ionic charge used as a mole ratio
  The $2+$ charge on the manganese ion was used as a stoichiometric factor and the amount divided by $2$ to $0.02\ \text{mol}$: $0.02 \times 252 = 5.04\ \text{kJ}$ and $5040 \div 840 = 6$. The charge counts the electrons transferred, not the atoms: each manganese ion gives one manganese atom, so the amount stays at $0.04\ \text{mol}$.
• D. 50.4 · Specific heat capacity omitted
  The specific heat capacity was dropped from the denominator: $10080 \div 200 = 50.4$. Dividing joules by a mass alone leaves $\text{J g}^{-1}$, an energy per gram and not a temperature at all; it is the $4.2$ that turns joules per gram into degrees.
• E. 300 · Scaling by the amount omitted
  The per-mole figure was treated as the total heat released: $252\ \text{kJ} = 252000\ \text{J}$ and $252000 \div 840 = 300$. Only $0.04\ \text{mol}$ of manganese is displaced, so only that fraction of the $252\ \text{kJ}$ is ever given out, and $300$ is the rise a whole mole would produce.

Common Mistake (⚠️):
Dividing the heat by the amount instead of multiplying, out of habit from the more familiar calorimetry question that asks for kilojoules per mole. Here the energy per mole is handed over, $252\ \text{kJ}$, and the temperature rise is what is wanted, so the amount $0.04\ \text{mol}$ multiplies up to a total heat of $10080\ \text{J}$ first, and only then does $Q = mc\Delta T$ get rearranged.

Takeaway (📌):
The Cheat Code: Work out what one degree costs before anything else. The product $mc$ is the price of a degree in joules, here $840\ \text{J}$, and every calorimetry question is then either a heat divided by that price or that price multiplied by a rise.

Correct Answer: B. 0.672

Fastest Approach (🚀):
$96\,\mathrm{cm^3}$ out of the $24000\,\mathrm{cm^3}$ in one mole is $0.004\,\mathrm{mol}$ of carbon dioxide; scale by $\tfrac{2}{1}$ to get $0.008\,\mathrm{mol}$, then multiply by $84$. Two options go without any arithmetic: whichever is a thousand times another is the $\mathrm{cm^3}$ against $\mathrm{dm^3}$ slip, and $0.008$ on its own is an amount in $\mathrm{mol}$, not a mass.

Step-by-Step Breakdown:

1. Recover the total volume of carbon dioxide

A mean rate is the total change divided by the time it took, so the total is the rate multiplied by the interval. Only the carbon dioxide reaches the syringe, so its reading follows that one substance:

2. Convert to moles of carbon dioxide, then to moles of sodium hydrogencarbonate

The balanced equation pairs $2\,\mathrm{NaHCO_3}$ with $1\,\mathrm{CO_2}$, so the amount of sodium hydrogencarbonate is $\tfrac{2}{1}$ times the amount of carbon dioxide:

3. Convert the amount to a mass

Check the units before committing. Dividing $96\,\mathrm{cm^3}$ by a molar volume quoted in $\mathrm{dm^3}$ would have given $4\,\mathrm{mol}$ of carbon dioxide and a final mass of $672\,\mathrm{g}$, which a syringe holding $96\,\mathrm{cm^3}$ cannot account for.

The key is $0.672$.

Why the Other Options Are Wrong (❌):

• A. 0.008 · Amount in moles reported as a mass
  The amount in moles was written down as though it were already a mass: $n(\mathrm{NaHCO_3})=0.008\,\mathrm{mol}$ was quoted as $0.008\,\mathrm{g}$. The last conversion is still owed, and it is the one the stem hands over: multiplying by $M_r(\mathrm{NaHCO_3})=84$ gives $0.672$.
• C. 0.336 · Stoichiometric ratio ignored
  The equation's coefficients were skipped and the amount of sodium hydrogencarbonate was set equal to the amount of carbon dioxide, $0.004\,\mathrm{mol}$ rather than $0.008\,\mathrm{mol}$, before multiplying by $84$. The equation pairs $2\,\mathrm{NaHCO_3}$ with $1\,\mathrm{CO_2}$, so the amount of sodium hydrogencarbonate is $\tfrac{2}{1}$ times the amount of carbon dioxide, and this option is the key $0.672$ scaled by $\tfrac{1}{2}$.
• D. 0.848 · Wrong relative formula mass used
  The right amount, $0.008\,\mathrm{mol}$, was multiplied by the relative formula mass of sodium carbonate instead: $106$ rather than the $84$ the stem supplies. $M_r(\mathrm{Na_2CO_3})=106$ is genuine, but it belongs to another substance in the equation, and the question asks for the mass of sodium hydrogencarbonate.
• E. 672 · Molar volume units mismatched
  The volume was left in $\mathrm{cm^3}$ and divided by a molar volume quoted in $\mathrm{dm^3}$: $96\div24=4\,\mathrm{mol}$ of carbon dioxide instead of $0.004\,\mathrm{mol}$. One mole of gas occupies $24\,\mathrm{dm^3}$, which is $24000\,\mathrm{cm^3}$, so this option is the key $0.672$ multiplied by a thousand.

Common Mistake (⚠️):
Starting the mole calculation from the rate itself. A figure in $\mathrm{cm^3\,s^{-1}}$ is not a volume, so it has to be multiplied by the $80\,\mathrm{s}$ interval before the molar gas volume can be used; feeding $1.2$ straight into the division gives an amount $80$ times too small.

Takeaway (📌):
The Cheat Code: Rate times time gives the amount of carbon dioxide; the molar gas volume, the $2$ to $1$ from the balanced equation and $M_r(\mathrm{NaHCO_3})=84$ then run the chain backwards to the sodium hydrogencarbonate. Following a product is only ever a way of following the reactant that made it.

Correct Answer: E. $208 \text{ kJ mol}^{-1}$

Fastest Approach (🚀):
Subtract $\Delta H$ from the forward barrier: $268 - (+60) = 208$. Matches Option E.

Step-by-Step Breakdown:

1. Picture the reaction profile

The forward activation energy is measured from the reactants up to the transition state; the reverse activation energy is measured from the products up to that same transition state.

2. Relate the two barriers

3. Substitute

4. Check it is physical

The reaction is endothermic ($\Delta H = +60$ kJ mol$^{-1}$), so the reverse barrier must be smaller than the forward one, and $208 < 268$. Both barriers are positive, as every activation energy must be.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $60 \text{ kJ mol}^{-1}$ · Confused quantities
  $60 \text{ kJ mol}^{-1}$ is the enthalpy change, the height difference between reactants and products. An activation energy is the height of the barrier above one of them, which is a different measurement on the same diagram.
• B. $416 \text{ kJ mol}^{-1}$ · Impossible on the reaction profile
  $416$ is larger than the forward barrier of $268$, which cannot happen when the products start higher than the reactants, and no combination of $268$ and $60$ produces it either.
• C. $268 \text{ kJ mol}^{-1}$ · Answered the wrong quantity
  $268 \text{ kJ mol}^{-1}$ is the forward activation energy, which the question supplies. The reverse barrier is the one being asked for.
• D. $328 \text{ kJ mol}^{-1}$ · Sign error
  $268 + 60 = 328$: $\Delta H$ added rather than subtracted. That would put the reverse barrier above the forward one, and for an endothermic reaction it has to be below.

Common Mistake (⚠️):
Adding the enthalpy change directly to the forward activation energy regardless of sign.

Takeaway (📌):
The Cheat Code: In an endothermic reaction the products sit above the reactants, so the climb from the product side up to the transition state is the shorter one: the reverse barrier must come out smaller than the forward barrier. That alone removes every option at or above $268 \text{ kJ mol}^{-1}$ before any arithmetic.

Correct Answer: B. 64

Fastest Approach (🚀):
Settle the charge step before touching the arithmetic, because the sign alone decides it. A positive ion has lost electrons and therefore holds more protons than electrons; a negative ion has gained them and holds fewer. Once the $30$ protons are on the page the rest is a single addition, and any option no larger than the $34$ neutrons can be struck out before it is done, since the protons take up the rest of the mass number.

Step-by-Step Breakdown:

1. Turn the electron count into the atomic number

Charge is the count of electrons missing or spare. This ion has lost two electrons relative to the neutral atom, so it carries two electrons fewer than protons:

$Z = 28 + 2 = 30$

That proton count is the atomic number, the figure written at the bottom left.

2. Add the neutrons to reach the mass number

The mass number totals protons and neutrons and nothing else, because an electron is far too light to register in it:

$A = Z + n = 30 + 34 = 64$

Sanity check: read the notation back. From $^{64}_{30}\mathrm{X}$ the neutrons are $64 - 30 = 34$, and an ion of charge $2+$ built on $30$ protons holds $30 - 2 = 28$ electrons. Both match what was measured.

The key is $64$.

Why the Other Options Are Wrong (❌):

• A. 34 · Neutron count read as the mass number
  Stops at the neutrons. The mass number counts protons and neutrons together, so $34$ omits every one of the $30$ protons; neutrons are never printed on their own in $^{A}_{Z}\mathrm{X}$, they are only ever the difference $A - Z$.
• C. 62 · Ion treated as a neutral atom
  Reads the $28$ electrons straight off as the proton count, which is true only of a neutral atom. This ion has lost two electrons, so the protons come to $28 + 2 = 30$ and the mass number to $30 + 34 = 64$.
• D. 60 · Charge applied in the wrong direction
  Moves the electron count the wrong way. A positive ion has lost two electrons, so it holds two electrons fewer than protons and the proton count is $28 + 2 = 30$, not $26$.
• E. 92 · Electrons counted into the mass number
  Adds the $28$ electrons to the nucleons as well. An electron is far too light to register in a mass number, which totals the $30$ protons and the $34$ neutrons and nothing else.

Common Mistake (⚠️):
Treating the ion as though it were neutral. Only in a neutral atom does the electron count double as the proton count; this ion has lost two electrons, so its $28$ electrons and its $30$ protons differ by $2$, and using $28$ as the atomic number shifts the mass number by that same $2$.

Takeaway (📌):
The Cheat Code: Charge first, then add. The charge turns the electron count into the proton count, $28 + 2 = 30$, and the mass number is that proton count plus the neutrons. Electrons never enter a mass number at all.

Correct Answer: A. 61

Fastest Approach (🚀):
Any option that is at least as large as the mass number $108$ can go immediately, because the protons must take up part of it. That removes two of the five before any subtraction is done.

Step-by-Step Breakdown:

1. Read the two numbers off the label, then take the difference

The label gives the nuclide as $^{108}_{47}\mathrm{Ag}$. The lower figure is the atomic number, $Z = 47$, and it counts protons only: it is what makes the sample silver rather than anything else. The upper figure is the mass number, $A = 108$, and it counts protons and neutrons together, because the electrons are far too light to register in it. Neutrons appear in $A$ and nowhere else on the label, so removing the protons from the mass number leaves them:

$n = A - Z = 108 - 47 = 61$

Sanity check: the neutron count cannot equal or exceed the mass number, since the protons take up part of it, so $108$ and $155$ are impossible before any arithmetic is attempted.

The key is $61$.

Why the Other Options Are Wrong (❌):

• B. 47 · Reads Z as the neutron count
  Quotes the atomic number instead of subtracting it. $Z = 47$ is the proton count, and the neutrons are what is left of the mass number once those protons are removed, $108 - 47$, so $47$ answers a different question.
• C. 108 · Reads A as the neutron count
  Reads the mass number as a neutron count. $A = 108$ is protons and neutrons together, so it already contains the $47$ protons and overstates the neutrons by exactly that many.
• D. 155 · Addition for subtraction
  Adds the two numbers, $108 + 47 = 155$, instead of subtracting them. That would make the nucleus heavier than the mass number printed on its own label.
• E. 14 · Subtracts Z twice
  Takes the protons off and then the electrons as well, $108 - 47 - 47 = 14$. The neutral atom does have $47$ electrons, but they sit outside the nucleus and were never counted in the mass number, so $Z$ comes off once only.

Common Mistake (⚠️):
Quoting the upper number as the neutron count. The mass number is protons and neutrons added together, not neutrons alone, so $108$ silently includes the $47$ protons and overstates the neutrons by exactly that many.

Takeaway (📌):
The Cheat Code: The bottom number identifies, the top number weighs. Protons come straight from the bottom number, neutrons only from the difference, and nothing on a nuclide label ever counts neutrons for you.