Question

In: Chemistry

Determine the equilibrium constant for the following reaction at 655 K. HCN(g) + 2 H2(g) →...

Determine the equilibrium constant for the following reaction at 655 K.

HCN(g) + 2 H2(g) → CH3NH2(g) ΔH° = -158 kJ; ΔS°= -219.9 J/K. Determine the equilibrium constant for 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

Solutions

Expert Solution

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 = e 2.56

K = 12.93 = 13

Answer is (B).

queen_honey_blossom answered 2 years ago
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