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 |
Answer:
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
Answer is (B).
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