In: Chemistry
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 | 
We know that
Go =
Ho -
T
So
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).