Question

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

calculate the equilibrium constant for this reaction at 298.15K.

Answer 1

Answer:

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