Consider the reaction C2H4 (g) + H2O (g) - -> CH3CH2OH (g)

calculate the equilibrium constant for this reaction at 298.15K.

K = 361.369

Explanation:

C2H4 (g) + H2O (g) → CH3CH2OH (g)

∴ ΔG°f (298.15K) CH3CH2OH (g) = - 174.8 KJ/mol

∴ ΔG°f (298.15) C2H4 (g) = 68.4 KJ/mol

∴ ΔG°f (298.15) H2O (g) = - 228.6 KJ/mol

⇒ ΔG°f (298.15) = - 174.8 - ( - 228.6 + 68.4) = - 14.6 KJ/mol

K = e∧ (-ΔG°f / RT)

∴ R = 8.314 E-3 KJ/mol. K

∴ T = 298.15 K

⇒ K = e∧ ( - (-14.6) / ((8.314 E-3) (298.15)))

⇒ K = e∧ (5.889)

⇒ K = 361.369