Question

Consider the following reaction and its equilibrium constant:
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)Kc=5.85\times {10}^{-3}$

A reaction mixture at a particular time, t contains: $\left[N{O}_2\right]=0.025Mand\left[N_2{O}_4\right]=0.033M.$ Calculate $Qc$, and determine the direction in which the reaction will proceed.

• A. $1.9\times {10}^{-3}$, the reaction will proceed in forward direction
• B. $1.9\times {10}^{-2}$, the reaction will proceed in backward direction
• C. $5.85\times {10}^{-3}$, it will be equal to the equilibrium constant
• D. None of the above

Answer 1

The correct option is B $1.9\times {10}^{-2}$, the reaction will proceed in backward direction

Given reaction is $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)Kc=5.85\times {10}^{-3}$

$\left[N{O}_2\right]=0.025M\left[N_2{O}_4\right]=0.033M$
Reaction Quotient $\left(Q_c\right)=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}$
$Q_c=\frac{(0.025{)}^2}{\left(0.033\right)}$

$Q_c\approx 0.019=1.9\times {10}^{-2}$

comparing $Q_c$ with $K_c$ we get $Q_c>K_c$
Hence, the reaction will proceed in backward direction.