IB Chemistry HL R1.4 Entropy and spontaneity questions

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Question 1

Multiple choiceMedium1 mark
2HOOH(l)HOOH(l)2HOH(l)HOH(l)+OO(g)hydrogen peroxide (SMILES: OO)water (SMILES: O)oxygen (SMILES: O=O)

For the reaction shown, the standard molar entropies are S°(H₂O₂(l)) = 109.6, S°(H₂O(l)) = 69.9 and S°(O₂(g)) = 205.0 J K⁻¹ mol⁻¹. Which option gives ΔᵣS° per mole of reaction as written and the resulting change in the system’s entropy?

Show answer and explanation

Answer: C. For a reaction, ΔᵣS° = ΣνS°(products) − ΣνS°(reactants). Using the coefficients read from the figure: [2(69.9) + 205.0] − [2(109.6)] = +125.6 J K⁻¹ mol⁻¹. The positive value means the system’s entropy increases.

A: reverses products minus reactants, giving −125.6 J K⁻¹ mol⁻¹ and a decrease.
B: omits the coefficient 2 for water, giving +55.7 J K⁻¹ mol⁻¹.
D: omits the coefficient 2 for hydrogen peroxide, giving +235.2 J K⁻¹ mol⁻¹.

Remember: multiply each standard molar entropy by its stoichiometric coefficient before summing.

Question 2

Paper 2Medium7 marks

Calcium carbonate decomposes on heating according to the reaction

\ceCaCO3(s)−>CaO(s)+CO2(g).\ce{CaCO3(s) -> CaO(s) + CO2(g)}.

The standard entropy values and standard enthalpy change for this reaction are given below. Figure 1 shows how ΔrG∘\Delta_\mathrm r G^\circ varies with temperature.

Substance or quantityValue
S∘(\ceCaCO3(s))S^\circ(\ce{CaCO3(s)})92.9 J K−1 mol−192.9\ \mathrm{J\,K^{-1}\,mol^{-1}}
S∘(\ceCaO(s))S^\circ(\ce{CaO(s)})39.8 J K−1 mol−139.8\ \mathrm{J\,K^{-1}\,mol^{-1}}
S∘(\ceCO2(g))S^\circ(\ce{CO2(g)})213.7 J K−1 mol−1213.7\ \mathrm{J\,K^{-1}\,mol^{-1}}
ΔrH∘\Delta_\mathrm r H^\circ+178.3 kJ mol−1+178.3\ \mathrm{kJ\,mol^{-1}}
8009001000110012001300-40-30-20-100102030405060Temperature, T / KΔr G^° / kJ mol-1
(a)

Calculate ΔrS∘\Delta_\mathrm r S^\circ for the reaction. Include the sign and units.

[2]
(b)

Using Figure 1, determine the approximate temperature at which ΔrG∘\Delta_\mathrm r G^\circ changes sign. State whether the forward reaction is thermodynamically favoured under standard-state conditions at 1000 K1000\ \mathrm K

[2]
(c)

Using your answer to (a), calculate ΔrG∘\Delta_\mathrm r G^\circ at 1200 K1200\ \mathrm K State whether the forward reaction is thermodynamically favoured under standard-state conditions at this temperature.

[3]

Question 3

Paper 2Hard12 marks

Calcium oxide is produced by heating limestone, which contains calcium carbonate. The standard Gibbs energy change, ΔrG∘\Delta_\mathrm r G^\circ for the decomposition is shown in Figure 1. Assume that ΔrH∘\Delta_\mathrm r H^\circ and ΔrS∘\Delta_\mathrm r S^\circ remain constant over the temperature range shown.

7008009001000110012001300-40-30-20-10010203040506070T / KΔr G° / kJ mol-1
(a)

Write a balanced equation, including state symbols, for the thermal decomposition of calcium carbonate.

[1]
(b)

Use Figure 1 to estimate ΔrG∘\Delta_\mathrm r G^\circ at 1000 K1000\ \mathrm K

[1]
(c)

Use Figure 1 to determine ΔrH∘\Delta_\mathrm r H^\circ and ΔrS∘\Delta_\mathrm r S^\circ for the decomposition. Give signs and units, and show how you obtain both values.

[3]
(d)

Using your values from (c), determine the temperature at which ΔrG∘=0\Delta_\mathrm r G^\circ=0 State at which temperatures the forward decomposition is spontaneous under standard-state conditions.

[2]
(e)

At 1000 K1000\ \mathrm K a gas stream keeps the carbon dioxide partial pressure in the kiln at 0.10 bar0.10\ \mathrm{bar} For the reaction as written, take

Q=p(CO2)p∘,p∘=1.00 bar,Q=\frac{p(\mathrm{CO_2})}{p^\circ},\qquad p^\circ=1.00\ \mathrm{bar},

with pure solids omitted from QQ Under these conditions,

ΔrG=ΔrG∘+RTln⁡Q,\Delta_\mathrm r G=\Delta_\mathrm r G^\circ+RT\ln Q,

where R=8.31 J K−1 mol−1R=8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}

Determine the carbon dioxide partial pressure at which ΔrG=0\Delta_\mathrm r G=0 at 1000 K1000\ \mathrm K Hence decide whether the forward decomposition is spontaneous at p(CO2)=0.10 barp(\mathrm{CO_2})=0.10\ \mathrm{bar} and assess the claim that a positive ΔrG∘\Delta_\mathrm r G^\circ at 1000 K1000\ \mathrm K means that the decomposition cannot be spontaneous.

[5]

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