Chemistry 9701 · AS & A Level · Chemical energetics

Chemical energetics — practice question

Calcium metal reacts with chlorine gas to produce calcium chloride, $\text{CaCl}_2$.
(a(i))[1]

Write an equation, including state symbols, that represents the lattice energy of calcium chloride, $\text{CaCl}_2$.

(a(ii))[2]

Complete a fully labelled Born-Haber cycle that could be used to determine the lattice energy, $\u000394H^\circ_{\text{latt}}$, for calcium chloride.

(a(iii))[3]

Use your answer to (ii) and the data below, together with the relevant data from the Data Booklet, to calculate a value for $\u000394H^\circ_{\text{latt}}$ for calcium chloride. Standard enthalpy change of formation of $\text{CaCl}_2(s)$, $\u000394H^\circ_f = -796\ \text{kJ mol}^{-1}$. Standard enthalpy change of atomisation of $\text{Ca}(s)$, $\u000394H^\circ_{\text{at}} = +178\ \text{kJ mol}^{-1}$. Electron affinity of chlorine atoms $= -349\ \text{kJ mol}^{-1}$. $\u000394H^\circ_{\text{latt}} = \ldots\ldots\ \text{kJ mol}^{-1}$

(b)[2]

Entropy is a measure of the disorder in a system. Describe and explain what happens to the entropy of a gas when the temperature increases.

(c(i))[2]

The table contains four reactions. For each one, state whether the entropy change, $\Delta S^\circ$, is positive, negative or, if there is no change in entropy, ‘no change’ in the table below. The first reaction has already been completed for you. $\text{CO}(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$ $\text{Mg}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{MgO}(s)$ $\text{CuSO}_4(s) + 5\text{H}_2\text{O}(l) \rightarrow \text{CuSO}_4\cdot5\text{H}_2\text{O}(s)$ $\text{NaHCO}_3(s) + \text{H}^+(aq) \rightarrow \text{Na}^+(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$

(c(ii))[1]

Explain why the entropy change for the first reaction is negative.

(d)[2]

Calculate the standard entropy change, $\Delta S^\circ$, for this reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ Standard entropies, $S^\circ$, in $\text{J K}^{-1}\text{mol}^{-1}$ are given: $\text{N}_2(g) = 192$, $\text{H}_2(g) = 131$, $\text{NH}_3(g) = 193$

(e(i))[2]

Whether or not a chemical reaction is spontaneous (feasible) can be decided by calculating the change in free energy, $\Delta G^\circ$, at a given temperature. For the reaction $\text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g)$ $\Delta H^\circ = +117\ \text{kJ mol}^{-1}$, $\Delta S^\circ = +175\ \text{J K}^{-1}\text{mol}^{-1}$ Calculate the value of $\Delta G^\circ$ at $298\ \text{K}$ for the reaction above.

(e(ii))[1]

Use your answer to (i) to explain whether or not this reaction is spontaneous at $298\ \text{K}$.

Worked solution & mark scheme

This 16-mark question has a full step-by-step worked solution and mark scheme. One marking point: Therefore, $Ca^{2+}(g) + 2Cl^-(g) \rightarrow CaCl_2(s)$, with the state symbols shown correctly

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