For the decomposition reaction ${NH}_{2} {COONH}_{4} (s) ⇌ 2 {NH}_{3} (g) + {CO}_{2} (g)$ , the $K_{p} = 2.9 \times 10^{- 5} {atm}^{3}$ . The total pressure of gases at equilibrium when $1.0 mol$ of ${NH}_{2} {COONH}_{4} (s)$ was taken to start with would be

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

0.0194 atm

0.0776 atm

0.0388 atm

0.0582 atm

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\underset{1.0 mol - x}{{NH}_{2} {COONH}_{4} (s)} \underset{2 p}{2 {NH}_{3} (g)} + \underset{p}{{CO}_{2} (g)}$

$K_{p} = p_{{NH}_{3}}^{2} p_{{CO}_{2}}$ i.e. $2.9 \times 10^{- 5} {atm}^{3} = (2 p)^{2} p$

This gives $p = {(\frac{2.9 \times 10^{- 5} atm}{4})}^{1 / 3} = 0.0194 atm$ and $p_{total} = 2 p + p = 3 p = 3 \times 0.0194 atm = 0.0582 atm$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE