Question 4 of 5: Temperature Dependence of ΔG° for Cr 3 C 2 Oxidation — Kirchhoff’s Law vs. the Constant-ΔH,ΔS Approximation

Question 4: Temperature Dependence of ΔG° for Cr₃C₂ Oxidation — Kirchhoff’s Law vs. the Constant-ΔH,ΔS Approximation (25%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The reaction 7Cr₃C₂(s) + 2.5O₂(g) → 3Cr₇C₃(s) + 5CO(g); C_p(T), ΔH°₂₉₈ and S°₂₉₈ for every species (all in cal, cal/mol·K).

Find. (a) ΔG°(T); (b) ΔG° at 1500 K and 298 K from the full Kirchhoff treatment; (c) the same two values from the constant-ΔH,ΔS shortcut; (d) the % difference between (b) and (c).

Approach. Sum each property over the reaction stoichiometry (products minus reactants) to get ΔH°₂₉₈, ΔS°₂₉₈ and ΔC_p(T); integrate Kirchhoff’s law to carry ΔH and ΔS from 298 K to T, then form ΔG(T) = ΔH(T) − TΔS(T).

1. ΔH°₂₉₈ and ΔS°₂₉₈ by Hess’s law (coefficients: +3 Cr₇C₃, +5 CO, −7 Cr₃C₂, −2.5 O₂): $$\Delta H^{\circ}_{298}=3(-54500)+5(-26420)-7(-26200)-2.5(0)=-112{,}200\text{ cal}$$ $$\Delta S^{\circ}_{298}=3(48.00)+5(47.22)-7(20.40)-2.5(49.00)=114.8\text{ cal/K}$$
2. ΔC_p(T) by the same stoichiometric sum of the four given C_p expressions: $$\Delta C_p(T)=-23.28+6.96\times10^{-3}T+\frac{22.31\times10^{5}}{T^2}\text{ cal/mol K}$$
3. Kirchhoff integration → ΔG°(T), part (a). $$\Delta H(T)=\Delta H^{\circ}_{298}+\Delta a(T-298)+\tfrac{\Delta b}{2}(T^2-298^2)-\Delta c\Big(\tfrac1T-\tfrac1{298}\Big)$$ $$\Delta S(T)=\Delta S^{\circ}_{298}+\Delta a\ln\tfrac{T}{298}+\Delta b(T-298)-\tfrac{\Delta c}{2}\Big(\tfrac1{T^2}-\tfrac1{298^2}\Big)$$ $$\boxed{\Delta G^{\circ}(T)=\Delta H(T)-T\,\Delta S(T)}\quad\text{with }\Delta a=-23.28,\ \Delta b=6.96\times10^{-3},\ \Delta c=22.31\times10^5$$
4. Part (b): evaluate the full expression. $$\Delta G^{\circ}(298\text{ K})=-146{,}410\text{ cal}\qquad \boxed{\Delta G^{\circ}(1500\text{ K})=-273{,}075\text{ cal}}$$ (the 298 K value is, as expected, identical to ΔH°₂₉₈−298ΔS°₂₉₈, since the Kirchhoff correction vanishes at the reference temperature).
5. Part (c): constant-ΔH,ΔS shortcut (ignores C_p entirely): $$\Delta G^{\circ}_{simple}(T)=\Delta H^{\circ}_{298}-T\,\Delta S^{\circ}_{298}$$ $$\Delta G^{\circ}_{simple}(298\text{ K})=-112{,}200-(298)(114.8)=-146{,}410\text{ cal}$$ $$\boxed{\Delta G^{\circ}_{simple}(1500\text{ K})=-112{,}200-(1500)(114.8)=-284{,}400\text{ cal}}$$
6. Part (d): percentage difference. At 298 K the two methods coincide exactly (0% difference, by construction). At 1500 K: $$\%\,\text{diff}=\frac{\big|\Delta G^{\circ}_{simple}-\Delta G^{\circ}\big|}{\big|\Delta G^{\circ}\big|}\times100 =\frac{|-284{,}400-(-273{,}075)|}{273{,}075}\times100$$ $$\boxed{\%\,\text{diff}(1500\text{ K})=4.15\%}$$