Question

Equilibrium constant, $K_C$ for the reaction
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ at $500K$ is $0.061$
At a particular time, the analysis shows that composition of the reaction mixture is $3.0mol{L}^{-1}{N}_2,2.0mol{L}^{-1}{H}_2$ and $0.5mol{L}^{-1}N{H}_3.$ Is the reaction at equilibirium and if not in which direction does the reaction tend to proceed to reach equilibrium?

• A. at equilibrium
• B. not at equilibrium, backwards shift
• C. not at equilibrium, forward shift
• D. can not be predicted

Answer 1

The correct option is C not at equilibrium, forward shift

$\begin{array}{cccccc}& N_2\left(g\right)& +& 3H_2\left(g\right)& \rightleftharpoons & 2N{H}_3\left(g\right)\\ Att& 3M& & 2M& & 0.5M\end{array}$
$Q=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$
$Q=\frac{0.5\times 0.5}{3\times 2^3}=\frac{1}{3\times 8\times 4}=\frac{1}{96}=0.0104$

$\because Q<K,$ the reaction will shift in forward direction.



Theory:

$K_c$ (equilibrium constant) is directly proportional to product of concentration of products each raised to the corresponding stoichiometric coefficients and inversely proportional to product of concentration of reactants raised to the corresponding stoichiometric coefficients.
When,
$K_c<{10}^{-3},$ the reaction will rarely proceed.
$K_c>{10}^{+3}$, the reaction will proceed towards completion.
${10}^{-3}<K_c<{10}^{+3}$, both the reactant and products are significant.