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Chemical Energetics: Question 10

Syllabus 5.1

Multiple choice AS 1 mark

A student adds 0.120 g0.120\ \text{g} of magnesium ribbon (molar mass 24.0 g mol−124.0\ \text{g mol}^{-1}) to 50.0 cm350.0\ \text{cm}^3 of 2.00 mol dm−32.00\ \text{mol dm}^{-3} hydrochloric acid, which is in excess, in an insulated cup:

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}

The temperature of the solution rises from 18.0 ∘C18.0\,^\circ\text{C} to 29.2 ∘C29.2\,^\circ\text{C}. Assume the mass of the solution remains 50.0 g50.0\ \text{g} and its specific heat capacity is 4.18 J g−1 K−14.18\ \text{J g}^{-1}\text{ K}^{-1}.

What is the enthalpy change of reaction, per mole of magnesium, for this reaction?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Identify the limiting reagent

Amount of Mg: n(Mg)=0.12024.0=0.00500 moln(\text{Mg}) = \frac{0.120}{24.0} = 0.00500\ \text{mol}

Amount of HCl present: n(HCl)=50.01000×2.00=0.100 moln(\text{HCl}) = \frac{50.0}{1000} \times 2.00 = 0.100\ \text{mol}

The equation requires 2 mol HCl2\ \text{mol HCl} per 1 mol Mg1\ \text{mol Mg}, so reacting all the magnesium would only need 2×0.00500=0.0100 mol2 \times 0.00500 = 0.0100\ \text{mol} of HCl. Since 0.100 mol0.100\ \text{mol} is available (far more than 0.0100 mol0.0100\ \text{mol}), HCl is in large excess and magnesium is the limiting reagent, all 0.00500 mol0.00500\ \text{mol} of Mg reacts.

Step 2: Calculate the heat energy released

The temperature rise is: ΔT=29.2−18.0=11.2 ∘C (=11.2 K)\Delta T = 29.2 - 18.0 = 11.2\,^\circ\text{C}\ (= 11.2\ \text{K})

Using q=mcΔTq = mc\Delta T: q=50.0×4.18×11.2q = 50.0 \times 4.18 \times 11.2

Working in two steps: 50.0×4.18=209 J K−150.0 \times 4.18 = 209\ \text{J K}^{-1} 209×11.2=2340.8 J209 \times 11.2 = 2340.8\ \text{J}

So q=2340.8 J=2.3408 kJq = 2340.8\ \text{J} = 2.3408\ \text{kJ}.

Check (independent recomputation): 209×11.2=209×11+209×0.2=2299+41.8=2340.8 J209\times11.2 = 209\times11 + 209\times0.2 = 2299+41.8=2340.8\ \text{J}, consistent.

Step 3: Calculate the enthalpy change per mole of magnesium

The heat calculated was released by reacting the limiting reagent, magnesium, so: ΔH=−qn(Mg)=−2.3408 kJ0.00500 mol\Delta H = -\frac{q}{n(\text{Mg})} = -\frac{2.3408\ \text{kJ}}{0.00500\ \text{mol}}

Dividing: 2.34080.00500=468.16\frac{2.3408}{0.00500} = 468.16

so ΔH=−468 kJ mol−1 (3 s.f.)\Delta H = -468\ \text{kJ mol}^{-1}\ (3\ \text{s.f.})

The value is negative because the temperature of the solution rose, showing the reaction released heat (exothermic).

Check (independent recomputation): 2340.8 J÷0.00500 mol=468 160 J mol−1=468.16 kJ mol−12340.8\ \text{J}\div0.00500\ \text{mol}=468\,160\ \text{J mol}^{-1}=468.16\ \text{kJ mol}^{-1}, confirming ΔH≈−468 kJ mol−1\Delta H\approx-468\ \text{kJ mol}^{-1}.

Why the other options are wrong

  • B (+468 kJ mol−1+468\ \text{kJ mol}^{-1}): correct magnitude but wrong sign, omits the negative sign despite the exothermic temperature rise.
  • C (−23.4 kJ mol−1-23.4\ \text{kJ mol}^{-1}): comes from dividing qq by the total moles of HCl in the flask (0.100 mol0.100\ \text{mol}) instead of the limiting reagent, Mg: 2340.8÷0.100=23 408 J mol−1=23.4 kJ mol−12340.8 \div 0.100 = 23\,408\ \text{J mol}^{-1} = 23.4\ \text{kJ mol}^{-1}.
  • D (−234 kJ mol−1-234\ \text{kJ mol}^{-1}): comes from dividing qq by the moles of HCl that actually reacted (0.0100 mol0.0100\ \text{mol}, from the 2:12:1 ratio) instead of moles of Mg reacted: 2340.8÷0.0100=234 080 J mol−1=234 kJ mol−12340.8 \div 0.0100 = 234\,080\ \text{J mol}^{-1} = 234\ \text{kJ mol}^{-1}.

Final answer

  • ΔH=−468 kJ mol−1\Delta H = \boxed{-468\ \text{kJ mol}^{-1}} (per mole of Mg), option A.