Question

GIven,1 mole nitrogen gas and 3 moles hydrogen gas form 0.05 moles $N{H_3}$ at equilibrium. What is ${K_c}$ of the reaction?

Answer 1

Hint: The equilibrium constant is denoted by ${K_c}$ . The letter c implies that the amount of reagents are given in molar concentrations. It is defined as the ratio of the concentration of products to the reactants raised to their respective stoichiometric. The value of ${K_c}$ for the reaction $A + B \rightleftharpoons AB$ can be given as: ${K_c} = \dfrac{{[AB]}}{{[A][B]}}$

Complete answer:
We are given the reaction of formation of ammonia from hydrogen and oxygen. The no. of moles of each reactant is given as 1 mole nitrogen gas and 3 moles hydrogen gas. The reaction can be given as:
${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$
The respective concentrations of each at equilibrium can be given by the following table. 1 mol of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia gas. The initial concentration of nitrogen and hydrogen can be considered as ‘a’. The amount used at equilibrium be ‘x’
${{\text{N}}_2}{\text{ + 3}}{{\text{H}}_3}{\text{ }} \rightleftharpoons {\text{ 2N}}{{\text{H}}_3}$

T=0	1	3	-
T=equilibrium	$1 - x$	$3 - 3x$	$2x$

We are given that at equilibrium the concentration of ammonia is 0.05M. Mathematically it can be given as: $2x = 0.05M \to x = \dfrac{{0.05}}{2} = 0.025M$
The concentrations of each species hence will be equal to: $[{N_2}] = 1 - x = 1 - 0.025 = 0.975M$
$[{H_2}] = 3 - 3x = 3 - 3(0.025) = 2.925M$
$[N{H_3}] = 0.05M$
The value of ${K_c}$ at equilibrium can be given as: ${K_c} = \dfrac{{{{[N{H_3}]}^2}}}{{[{N_2}]{{[{H_2}]}^3}}}$
${K_c} = \dfrac{{{{[0.05]}^2}}}{{[0.975]{{[2.925]}^3}}}$
${K_c} = 1.02 \times {10^{ - 4}}mo{l^{ - 1}}L$
This is the required answer.

Note:
Equilibrium constant ${K_c}$ doesn’t have any units, hence it is unitless if the no. of moles of reactants is equal to that of products. In here the no. of moles of reactant and products are different. Cancelling off the mol/L on the numerator and denominator we are left off with L/mol ($mo{l^{ - 1}}L$) . The concentration can be given as $c = \dfrac{n}{V}$ , but if the volume isn’t given consider it as 1L.