# Determine the equilibrium cons

Determine the equilibrium constant for the following reaction at655 K.

HCN(g) + 2 H2(g) → CH3NH2(g)ΔH° = -158 kJ; ΔS°= -219.9 J/K. Determine the equilibrium constantfor the following reaction at 655 K.HCN(g) + 2 H2(g) → CH3NH2(g) ΔH° =-158 kJ; ΔS°= -219.9 J/K

 A. 3.07 × 1011 B.13.0 C. 3.26 × 10-12 D. 3.99 × 1012 E. 2.51 × 10-13

We know that

Go =Ho -TSo

Given:

Ho =-158 kJ = – 158000 J

So =-219.9 J/K

T = 655 K

So,

Go =(- 158000 J) – (655 K) (-219.9 J/K)

= – 158000 J + 144034.5 J

= – 13965.5 J

Again,

Go = -RT lnK

Where K is the equilibirium constant.

So,

– 13965.5 J = – (8.314 J K-1mol-1) (655 K)ln K

13965.5 =5445.67 ln K

ln K =13965.5 / 5445.67

ln K =2.56

K = e2.56

K = 12.93 =13

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