Question

Ammonia dissociates into N2 and H2 such that degree of dissociation a is very less than 1 and equilibrium pressure is P0 then the value of a is [if Kp for 2NH3(g) $\rightleftharpoons$ N2 (g) + 3H2(g) is 27 × 10–8 P02] :

• A. 10–4
• B. 4 × 10–4
• C. 0.02
• D. Can’t be calculated.

Answer 1

Answer: (C)

0.02

Answer 2

2NH3(g) N2(g) + 3H2(g)

$1 - \alpha$ $\frac { \alpha } { 2 }$ $\frac { 3 \alpha } { 2 }$

$1 - \alpha$ 1 and $1 + \alpha$ 1

$\alpha = 2 \times 10 ^ { - 2 }$