From these data, calculate the value of ΔH° for the reaction.

Question

The equilibrium constant for the reaction below decreases from 11160.0 at 520.0°C to 17.0 at 820.0°C.

4NO + 3O₂ ---> 2N₂O₅

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Answer 1

K_eq = e^–ΔG°/RT means that ln(K_eq) = –ΔG°/RT

Since ΔG° = ΔH° – TΔS°, this means

ln(K_eq) = –ΔH°/RT + ΔS°/R

which we can rewrite as

ln(K_eq) = (–ΔH°/R)(1/T) + ΔS°/R

A plot of ln(K_eq) vs 1/T (called a Van't Hoff plot, BTW) gives a line with

slope = –ΔH°/R °K and

y-intercept = ΔS°/R

T₁ = (520 + 273) = 793 °K, K_eq1 = 11160

T₂ = (820 + 273) = 1093 °K, K_eq2 = 17

slope = [ln(K_eq2) – ln(K_eq1)] / [1/T₂ – 1/T₁] = [ln(17) – ln(11160)] / [1/1093 – 1/793] ≈ 18,742 = –ΔH°/R

ΔH° ≈ –18,742 × R = –18,742 °K × 8.314 J/mol-°K = –155,821 J/mol = –155.821 kJ/mol

Answer (rounded to 4 significant figures)

ΔH° = –155.8 kJ/mol