Question

At 700 K and 350 bar, a 1:3 mixture of ${N_2}\left( g \right)\;$ and ${H_2}\left( g \right)$ reacts to form an equilibrium mixture containing X_\left( {N{H_3}} \right) = 0.50. Assuming ideal behaviour.
A. {K_p} = 2.03 \times {10^{ - 4}} \\\
B. {K_p} = 3.55 \times {10^3} \\\
C. {K_p} = 3.1 \times {10^{ - 4}} \\\
D. $X_{N_2} = 0.125$

Answer 1

Equilibrium refers to a condition when the rate of forward reaction is equal to the rate of reverse reaction. The equilibrium constant, denoted by K, expresses the relationship between reactants and products of a reaction at an equilibrium condition with respect to a specific unit.

Complete answer:
For a generalised chemical reaction taking place in a solution:
$aA + bB \rightleftharpoons cC + dD\;$
The equilibrium constant can be expressed as follows:
$K = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
where [A], [B], [C] and [D] refer to the molar concentration of species A, B, C, D respectively at equilibrium. The coefficients like a, b, c, and d in the generalised chemical equation become exponents as seen in the above expression.
In the question, it is given that a 1:3 mixture of ${N_2}\left( g \right)\;$and ${H_2}\left( g \right)$reacts to form an equilibrium mixture. The chemical equation for this can be written as follows:
${N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right)$

| $N_2(g)$| $H_2(g)$| $NH_3(g)$
---|---|---|---
Initial No. of Moles| 1 | 3| 0
No. of moles at equilibrium| 1-x| 3-3x| 2x

Total number of moles of the chemical equation$= 1 - x + 3 - 3x + 2x = \left( {4 - 2x} \right)$
X_\left( {N{H_3}} \right) = 0.5(Given)
This means that:
$\Rightarrow \dfrac{2x}{4-2x} = 0.5 \\\ \Rightarrow x = \dfrac{2}{3} = 0.66$
Now, at equilibrium:

$$Moles{\text{ }}of\;{N_2} = 1 - \dfrac{2}{3} = \dfrac{1}{3} \\\ Moles{\text{ }}of\;{H_2} = 3 - 3 \times \dfrac{2}{3} = 1 \\\ {Moles{\text{ }}of\;N{H_3} = 2 \times \dfrac{2}{3} = \dfrac{4}{3}}$$

Now. Substituting all the values in the formula of equilibrium constant:
$\Rightarrow {K_P} = \dfrac{{{{[N{H_3}]}^2}}}{{[{N_2}]{{[{H_2}]}^3}}} = \dfrac{{{{\left( {\dfrac{4}{3}} \right)}^2}}}{{\dfrac{1}{3} \times {{\left( 1 \right)}^3}}} = \dfrac{{16}}{3} = 5.33 \\\ \Rightarrow X_{({N_2})} = \dfrac{{1 - x}}{{4 - 2x}} = \dfrac{{\dfrac{1}{3}}}{{4 - \dfrac{4}{3}}} = \dfrac{1}{{12 - 4}} = \dfrac{1}{8} = 0.125$

**Hence, the correct answer is Option D.

Note:**
While calculating the value of K, always remember few points such as
(i) K is constant for a particular reaction at a particular temperature. K changes on changing the temperature,
(ii) Pure solids or pure liquids are not generally included in the expression of equilibrium and
(iii) The chemical reaction must always be balanced including the coefficients (lowest possible integer values) to attain the correct value for K.