CAIE A-Level Chemistry 23.4 Gibbs Free Energy Change

Practise using ΔG° = ΔH° − TΔS° to calculate feasibility and predict how temperature changes it.

Exam points

convert ΔS from J K−1 mol−1 to kJ K−1 mol−1 before combining it with ΔH
use absolute temperature in kelvin and retain the signs of ΔH and ΔS throughout
interpret negative ΔG as thermodynamic feasibility without claiming that the reaction is fast

Covered syllabus points

23.4.1—The Gibbs equation ΔG 23.4.2—Calculations: the equation ΔG 23.4.3—Feasibility from ΔG sign 23.4.4—Temperature effect on feasibility

[Maximum number: 2]

(a)

For a particular gas phase reaction the variation in standard Gibbs free energy change, $\Delta G^{\ominus}$ , with temperature is shown.

Assume standard enthalpy change, $\Delta H^{\ominus}$ , and standard entropy change, $\Delta S^{\ominus}$ , remain constant with temperature.

Figure for Question (a) — CAIE A-Level Chemistry A2

[ 2 ]

(i)

Write the equation that relates $\Delta G^{\ominus}$ to $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ .

[ 1 ]

$\Delta G^{\ominus} = \Delta H^{\ominus} - \mathrm{T}\Delta S^{\ominus}$

(ii)

Use this equation to explain why $\Delta G^{\ominus}$ becomes less positive as temperature increases in this reaction.

[ 1 ]

$\mathrm{T} \Delta S$ is more positive

OR -T $\Delta \mathrm{S}$ becomes more negative [1]

[Maximum number: 3]

(a)

Whether or not a chemical reaction is spontaneous (feasible) can be deduced by calculating the change in free energy, $\Delta G^{\ominus}$ , at a given temperature.

\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})

\begin{aligned} & \Delta H^{\ominus}=+117 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta S^{\ominus}=+175 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}

[ 3 ]

(i)

Calculate the value of $\Delta G^{\ominus}$ at 298 K for the above reaction.

[ 2 ]

$\begin{aligned} \Delta G^{\ominus}=\Delta H^{\ominus}-\mathrm{T} \Delta S^{\ominus} =117-((298 \times 175) / 1000) =(+) 64.85\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right) \end{aligned}$

(ii)

Use your answer to (i) to explain whether or not this reaction is spontaneous at 298 K .

[ 1 ]

$\Delta G^{\ominus}$ is positive and so the reaction is not spontaneous (at 298 K )

[Maximum number: 4]

Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry.

(a)

KI forms an ionic lattice that is soluble in water.

[ 2 ]

(i)

KI(s) has a high solubility in water although its enthalpy change of solution is endothermic.

Explain how this high solubility is possible.

[ 2 ]

there is a (large) increase in entropy $\mathbf{OR} \Delta \mathrm{S}$ is positive $\mathbf{OR} \mathrm{T} \Delta \mathrm{S}$ is positive

so $\Delta G$ is negative / $T \Delta S$ outweighs $\Delta H$

(b)

KI slowly oxidises in air, forming $\mathrm{I}_{2}$ .

$reaction $1 \quad 4 \mathrm{KI}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}(\mathrm{~s})+2 \mathrm{I}_{2}(\mathrm{~s}) \quad \Delta H^{\ominus}=-203.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

Table 1.2 shows some data relevant to this question.

Table 1.2

[ 2 ]

(i)

Use your answer to (c)(i) to show that reaction 1 is spontaneous at 298 K .

[ 2 ]

$\Delta G=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ AND use of 298 K for T OR clear working to represent this

$\Delta G=-203.4-298(-514.4 / 1000)$ $\Delta G=\mathbf{- 5 0 . 1}$ OR $\mathbf{- 5 0 . 2}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ OR $\Delta G=\mathbf{- 5 0 1 0 8 . 8 ~ J ~ m i n} 3 \mathrm{sf}$ ECF from (c)(i)

(negative so spontaneous)

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Covered syllabus points

23.4 Gibbs free energy change, ΔG question 1

Question (a)

Question (i)

Question (ii)

23.4 Gibbs free energy change, ΔG question 2

Question (a)

Question (i)

Question (ii)

23.4 Gibbs free energy change, ΔG question 3

Question (a)

Question (i)

Question (b)

Question (i)