Question

Consider following reaction in equilibrium with equilibrium concentration 0.01 M of every species
(I) $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
(II) $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
(III) $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
Extend of the reaction taking place is:

• A. $I>II>III$
• B. $I<II<III$
• C. $II<III<I$
• D. $III<I<II$

Answer 1

The correct option is B $I<II<III$

The equilibrium constant of a chemical reaction is the value of the reaction quotient when the reaction has reached equilibrium.For a general chemical equilibrium,$\alpha A+\beta B\rightleftharpoons \rho R+\sigma S$,$K^{}c=\frac{{\left[R\right]}^{\rho }{\left[S\right]}^{\sigma }}{{\left[A\right]}^{\alpha }{\left[B\right]}^{\beta }}$. Here the concentration of reactants and products are taken instead of activities.

For comparing the extent of reaction of different equilibrium reactions,the value of $K_c$ is to be calculated. Greater the value of $K_c$,more is the extent of reaction.This comes from the Gibbs free energy relation, ${\Delta }_r{G}^{\circ }=-RT\left(\ln K_c\right)=-2.303RT\left({\log }_{10}{K}_c\right)$, more is the $K_c$,more is the $\Delta G^{\circ }$ and hence greater the extend of reaction.

Coming back to the question, $K_c$ of I is $K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$=$\frac{0.01\times 0.01}{0.01}$=0.01

$K_c$ of II is $K_c=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}$ is $\frac{0.01\times 0.01}{{0.01}^2}$=1

$K_c$ of III is $K_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$=$\frac{{0.01}^2}{0.01{0.01}^3}$=${10}^4$

Hence $K_c$'s of I < II < III