Question

Using the following data determine the temperature (in K) at which the reaction H₂O(g)+ C(s,graphite) ↔ H₂(g) + CO(g) this becomes spontaneous.
Δ_fH° (H₂O(g)) = -251.2 kJ mol^-1
Δ_fH° (C(s,graphite)) = 0.0 kJ mol^-1
Δ_fH° (H₂(g)) = 0.0 kJ mol^-1
Δ_fH° (CO(g)) = -110.1 kJ mol^-1
S° (H₂O(g)) = 192.6 J K^-1 mol^-1
S° (C(s,graphite)) = 6.4 J K^-1 mol^-1
S° (H₂(g)) = 136.9 J K^-1 mol^-1
S° (CO(g)) = 192.4 J K^-1 mol^-1

Answer 1

H₂O + C -> H₂fco (g) cg raphite) 695 cos shee CH2@g) = -251.2kJ/mol It's Egraphite) = 000 kJ/mole shoes Boots = 0.0 ks hole. *** C = -110.1K5 /mole 1 AH = AHf products - Alt of reactent shyz -110.1 +251.2 DHg =14101k5l mole S" (H2O) =. 192.6 Jkt mot! s graphite) = 604 J K molt So try = 136:9 J K molt 5 co = 192.4.5 k moet ... 69) 8 = 8 Products - Reactente

S-13609+192.4 –198.6–604 8 = 130.3 I kl molt At Equilibrium AG =0 m ДН = ТО, Tasha 14101x103 J. mort AS 130.3 J kl moly T= 1.0828 x103 kelvin T=1082. 8k the reaction is 1083t above spontanious.

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