Magnesium carbonate decomposes on heating: MgCO3(s) → MgO(s) + CO2(g) For this reaction, H^ = +117 kJ mol^-1. The table gives the standard entropy of each substance. Substance S^ / J K^-1mol^-1 MgCO3(s) 65.1 MgO(s) 26.9 CO2(g) 213.6 (a) Calculate the standard entropy change, S^,…

Answer: ((a)) S^ = +175.4 J K^-1mol^-1 ((b)) Positive, because one mole of solid reactant forms one mole of gas among the products; gas particles are far more disordered (many more accessible microstates) than solid particles, so the total entropy of the system increases. ((c)) T = 667 K (3 s.f.) ((d)) At 293 K, G^ = +65.6 kJ mol^-1 (positive), so the reaction is not thermodynamically feasible at this temperature.. Part (a): Standard entropy change Using S^ = S^ (products) - S^ (reactants): S^ = [S^ (MgO) + S^ (CO2)] - S^ (MgCO3) S^ = (26.9 + 213.6) - 65.1 = 240.5 - 65.1 = 175.4 J K^-1mol^-1 Check (independent recomputation): 26.9+213.6=240.5; 240.5-…

A2 Chemistry 9701: Entropy, Gibbs Free Energy and Reaction Feasibility

Substance	$S^{\ominus}$ / $\text{J K}^{-1}\text{mol}^{-1}$
$\text{MgCO}_3\text{(s)}$	$65.1$
$\text{MgO(s)}$	$26.9$
$\text{CO}_2\text{(g)}$	$213.6$

Substance

S^{\ominus}

\text{J K}^{-1}\text{mol}^{-1}

\text{MgCO}_3\text{(s)}

65.1

\text{MgO(s)}

26.9

\text{CO}_2\text{(g)}

213.6

Part (a): Standard entropy change

Using $\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})$ :

$\Delta S^{\ominus} = \left[S^{\ominus}(\text{MgO}) + S^{\ominus}(\text{CO}_2)\right] - S^{\ominus}(\text{MgCO}_3)$

$\Delta S^{\ominus} = (26.9 + 213.6) - 65.1 = 240.5 - 65.1 = 175.4\ \text{J K}^{-1}\text{mol}^{-1}$

Check (independent recomputation): $26.9+213.6=240.5$ ; $240.5-65.1=175.4$ , consistent.

Part (b): Why $\Delta S^{\ominus}$ is positive

The reaction converts one mole of solid into one mole of solid plus one mole of gas. Gas particles move freely and can occupy far more positions and energy states than particles fixed in a solid lattice, so the number of ways of arranging the particles and their energy (the number of accessible microstates) increases sharply. This increase in disorder means the total entropy of the system rises, giving a positive $\Delta S^{\ominus}$ .

Part (c): Minimum temperature for feasibility

A reaction becomes just feasible when $\Delta G^{\ominus} = 0$ , i.e. when: $\Delta H^{\ominus} = T\Delta S^{\ominus}$

Rearranging for $T$ : $T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}$

$\Delta H^{\ominus}$ is given in $\text{kJ mol}^{-1}$ , so $\Delta S^{\ominus}$ must be converted to the same energy unit: $\Delta S^{\ominus} = 175.4\ \text{J K}^{-1}\text{mol}^{-1} = 0.1754\ \text{kJ K}^{-1}\text{mol}^{-1}$

Substituting: $T = \frac{117}{0.1754} = 667\ \text{K} \ (3\ \text{s.f.})$

Check (independent recomputation): $0.1754 \times 667 = 0.1754\times600 + 0.1754\times67 = 105.24+11.75=116.99 \approx 117$ , consistent, confirming $T \approx 667\ \text{K}$ .

Above this temperature, $T\Delta S^{\ominus}$ exceeds $\Delta H^{\ominus}$ , making $\Delta G^{\ominus}$ negative and the decomposition feasible.

Part (d): Feasibility at 293 K

Using $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ at $T = 293\ \text{K}$ :

$T\Delta S^{\ominus} = 293 \times 0.1754 = 51.4\ \text{kJ mol}^{-1}$

$\Delta G^{\ominus} = 117 - 51.4 = +65.6\ \text{kJ mol}^{-1}$

Check (independent recomputation): $293\times0.1754 = 293\times0.17+293\times0.0054 = 49.81+1.58=51.39$ ; $117-51.39=65.61\approx65.6$ , consistent.

Since $\Delta G^{\ominus}$ is positive at $293\ \text{K}$ (well below the $667\ \text{K}$ minimum found in (c)), the reaction is not thermodynamically feasible at this temperature.

Final answers

(a) $\Delta S^{\ominus} = \boxed{+175.4\ \text{J K}^{-1}\text{mol}^{-1}}$
(b) Positive because a solid reactant produces a gaseous product, which is far more disordered than a solid.
(c) Minimum feasibility temperature $= \boxed{667\ \text{K}}$ (3 s.f.)
(d) $\Delta G^{\ominus}(293\ \text{K}) = \boxed{+65.6\ \text{kJ mol}^{-1}}$ , not feasible at 293 K.