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Using the following data determine the temperature (in K) at which the reaction H2O(g)+ C(s,graphite) ↔...

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.
ΔfH° (H2O(g)) = -251.2 kJ mol-1
ΔfH° (C(s,graphite)) = 0.0 kJ mol-1
ΔfH° (H2(g)) = 0.0 kJ mol-1
ΔfH° (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

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Answer #1

H₂O + C -> H₂fco (g) cg raphite) 695 cos shee CH2@g) = -251.2kJ/mol Its Egraphite) = 000 kJ/mole shoes Boots = 0.0 ks hole.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.

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