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  4. N 2+3 H 2 leftharpoons 2 NH 3 ; K c =1.2 At the start of a reaction, there are 0.249 mol N 2, 3.21 × 10-2 mol H 2 and 6.42 × 10-4 mol NH 3 in a 3.50 L reaction vessel at 375° C. Hence reaction will proceed in

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Q. $N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3} ; K _{ c }=1.2$
At the start of a reaction, there are $0.249\, mol \,N _{2}, 3.21 \times 10^{-2} mol\, H _{2}$ and $6.42 \times 10^{-4} mol \,NH _{3}$ in a $3.50 \,L$ reaction vessel at $375^{\circ} C$. Hence reaction will proceed in

Equilibrium

Solution:

$Q_{C}=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}+\frac{\left(\frac{6.42}{3.5}\right)^{2} \times\left(10^{-4}\right)^{2}}{\left(\frac{0.249}{3.5}\right)\left(\frac{3.21 \times 10^{-2}}{3.5}\right)^{3}}=0.61$
$\Rightarrow Q_{C} < K_{c}$
$ \Rightarrow $ Reaction proceeds in forward direction.