1. Tardigrade
  2. Question
  3. Chemistry
  4. For the reaction NH 2 COONH 4( s ) leftharpoons 2 NH 3( g )+ CO 2( g ) the equilibrium constant KP=2.92 × 10-5 atm 3 . The total pressure of the gaseous products when 1 mol of reactant is heated will be:

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Q. For the reaction
$NH _{2} COONH _{4}( s ) \rightleftharpoons 2 NH _{3}( g )+ CO _{2}( g )$
the equilibrium constant $K_{P}=2.92 \times 10^{-5}\, atm ^{3} .$ The total pressure of the gaseous products when $1\, mol$ of reactant is heated will be:

Equilibrium

Solution:

$\underset{(1-x)}{NH _{2} COONH _{4}(s)} \rightleftharpoons \underset{2x}{2 NH _{3}(g)}+ \underset{x}{CO _{2}(g)}$

Only gaseous substance are $NH _{3}$ and $CO _{2}$

$\therefore $ Total mol of gaseous substances $=3 x$

$\therefore p_{ NH _{3}} =p_{eq} \cdot \frac{2 x}{3 x}=\frac{2}{3} p_{ eq }$

$p_{ CO _{2}}=\frac{1}{3} p_{ eq },\,\,\, K_{p}=\left(p_{ NH _{3}}\right)^{2}\left(p_{ CO _{2}}\right)$

$=\left(\frac{2}{3} p_{ eq }\right)^{2}\left(\frac{1}{3} p_{ eq }\right)=2.92 \times 10^{-5}$

$\therefore p_{ eq }= 0.0582$ atm