1. Tardigrade
  2. Question
  3. Chemistry
  4. Ammonium carbamate decomposes as, NH 2 COONH 4(s) leftharpoons 2 NH 3(g)+ CO 2(g) For the reaction, Kp=2.9 × 10-5 atm 3 If we start with 1 mole of the compound, the total pressure at equilibrium would be

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Q. Ammonium carbamate decomposes as,
$NH _{2} COONH _{4(s)} \rightleftharpoons 2 NH _{3(g)}+ CO _{2(g)}$
For the reaction, $K_{p}=2.9 \times 10^{-5} \,atm ^{3} $ If we start with $1$ mole of the compound, the total pressure at equilibrium would be

Equilibrium

Solution:

$NH _{2} COONH _{4(s)} \rightleftharpoons 2 NH _{3(g)}+ CO _{2( g )}$

At eqm, if partial pressure of $CO _{2}=p$, then that of $NH _{3}=2 p$.

$K_{p}=p_{ NH _{3}}^{2} \times p_{ CO _{2}}=(2 p)^{2} \times p=4 p^{3}$

$2.9 \times 10^{-5}=4 p^{3}$ or $p^{3}=7.25 \times 10^{-6} $

$\Rightarrow p=1.935 \times 10^{-2}$

Hence, total pressure $=3 p=5.81 \times 10^{-2}=0.0581 \,atm$