Calculating ΔG° from ΔH° and ΔS° What is the standard free-energy change, ΔG°, for the following reaction at 25°C? N2(g) + 3H2(g) → 2NH3(g) Use values of ΔH°f and S°, from Tables 6.2 and 19.1. PROBLEM STRATEGY Calculate ΔH° and ΔS°, then substitute these values into ΔH° TΔS° to obtain ΔG°.

Question

Calculating ΔG° from ΔH° and ΔS°

What is the standard free-energy change, [latex]\Delta G^{\circ}[/latex], for the following reaction at [latex]25^{\circ} \mathrm{C}[/latex] ?

[latex]\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)[/latex]

Use values of [latex]\Delta H_{f}^{\circ}[/latex] and [latex]S^{\circ}[/latex], from Tables 6.2 and 19.1 .

PROBLEM STRATEGY

Calculate [latex]\Delta H^{\circ}[/latex] and [latex]\Delta S^{\circ}[/latex], then substitute these values into [latex]\Delta H^{\circ}-T \Delta S^{\circ}[/latex] to obtain [latex]\Delta G^{\circ}[/latex].

Accepted Answer

Write the balanced equation and place below each formula the values of [latex]\Delta H_{f}^{\circ}[/latex] and [latex]S^{\circ}[/latex] multiplied by stoichiometric coefficients.

[latex] \begin{array}{cccc} \mathrm{N}_{2}(g)+ & 3 \mathrm{H}_{2}(g) & & 2 \mathrm{NH}_{3}(g) \ \Delta H_{f}^{\circ}: 0 & 0 & & 2 imes(-45.9) \mathrm{kJ} \ S^{\circ}: 191.6 & 3 imes 130.6 & & 2 imes 192.7 \mathrm{~J} / \mathrm{K} \end{array} [/latex]

You calculate [latex]\Delta H^{\circ}[/latex] and [latex]\Delta S^{\circ}[/latex] by taking values for products and subtracting values for reactants.

[latex] \begin{aligned} \Delta H^{\circ} & =\sum n \Delta H_{f}^{\circ}( ext { products })-\sum m \Delta H_{f}^{\circ}( ext { reactants }) \ & =[2 imes(-45.9)-0] \mathrm{kJ}=-91.8 \mathrm{~kJ} \ \Delta S^{\circ} & =\sum n S^{\circ}( ext { products })-\sum m S^{\circ}( ext { reactants }) \ & =[2 imes 192.7-(191.6+3 imes 130.6)] \mathrm{J} / \mathrm{K}=-198.0 \mathrm{~J} / \mathrm{K} \end{aligned} [/latex]

You now substitute into the equation for [latex]\Delta G^{\circ}[/latex] in terms of [latex]\Delta H^{\circ}[/latex] and [latex]\Delta S^{\circ}[/latex]. Note that you substitute [latex]\Delta S^{\circ}[/latex] in units of [latex]\mathrm{kJ} / \mathrm{K}[/latex].

[latex]\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=-91.8 \mathrm{~kJ}-(298 \mathrm{~K})(-0.1980 \mathrm{~kJ} / \mathrm{K})=-\mathbf{3 2 . 8} \mathbf{k J}[/latex]

Question 19.4: Calculating ΔG° from ΔH° and ΔS° What is the standard free-e...

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