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

G^o = H^o - TS^o

Given:

H^o = -158 kJ = - 158000 J

S^o = -219.9 J/K

T = 655 K

So,

G^o = (- 158000 J) - (655 K) (-219.9 J/K)

= - 158000 J + 144034.5 J

= - 13965.5 J

Again,

G^o = - 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 1 year ago

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