Question

Equilibrium constant $K$ for the following reaction of $800K$ is $4$.
$N{H_3}(g)\underset {} \leftrightarrows \dfrac{1}{2}{N_2}(g) + \dfrac{3}{2}{H_2}(g)$
The value of $K$ for the following reaction will be :
${N_2}(g) + 3{H_2}(g)\underset {} \leftrightarrows 2N{H_3}(g)$
A.${(800\dfrac{R}{4})^{ - 2}}$
B.$16 \times {(800R)^2}$
C.${[\dfrac{1}{{4 \times 800R}}]^2}$
D.${(800R)^{10}} \times 4$

Answer 1

Hint: The second that has been given is clearly the reverse of the above reaction with coefficients of reactants and products changed. We know the equilibrium constant for reverse reaction is reciprocal of the forward reaction. We will use this relation to find the equilibrium constant for the second reaction.

Complete Step-by-step Solution:
First of all, let us study what equilibrium constant means in a chemical reaction.
Consider a general chemical reaction at equilibrium:
$aA + bB\underset {} \leftrightarrows cC + dD$
$
  K = \dfrac{{[products]}}{{[reactnt]}} \\
\Rightarrow K = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}} \\
 $

Where $K = \dfrac{[}{2}$$[A],[B],[C],[D]$ are the concentrations of $A,B,C,D$ at equilibrium.
 And $a,b,c,d$ are the number of moles of each of them respectively.

The reaction that is given to us :
$N{H_3}\underset {} \leftrightarrows \dfrac{1}{2}{N_2} + \dfrac{3}{2}{H_2}$ ……..(1)
Let us call this a reference reaction.
Equilibrium constant of above reaction is given:
$K = \dfrac{{{{[{N_2}]}^{\dfrac{1}{2}}}{{[{H_2}]}^{\dfrac{3}{2}}}}}{{[N{H_3}]}}$
$\Rightarrow$ $K = 4$
The backward reaction of eq(1) will be-
$\dfrac{1}{2}{N_2} + \dfrac{3}{2}{H_2}\underset {} \leftrightarrows N{H_3}$ ……(2)
The equilibrium constant for backward reaction is reciprocal of the forward reaction . So equilibrium constant for this reaction will be:
$
  {K_{backward}} = \dfrac{1}{{{K_{forward}}}} \\
\Rightarrow K' = \dfrac{1}{K} \\
 $

The reaction that has been given to us is :
${N_2} + 3{H_2}\underset {} \leftrightarrows 2N{H_3}$
This is the twice of the reaction in eq(2), value of equilibrium constant :
$K'' = {(K')^2} = {(\dfrac{1}{K})^2}$ …….(3)

We have derived the relation by which we are going to find its value:
Now in the reference equation all the elements are in gaseous states.
So here, the constant that is given is ${K_c}$.
We know that: ${K_p} = {K_c}{(RT)^{\Delta n}}$
$\Delta n = $number of moles of product-number of moles of reactants.
${K_p}$ for eq(1) is equal to –
${K_p} = 4(800R)$
${K_p}'$ for the asked reaction will be-
$
  {K_p}'' = \dfrac{1}{{{K_p}}} \\
\Rightarrow {K_p}'' = \dfrac{1}{{{{(4 \times 800R)}^2}}} \\
 $

Hence, the correct option is C.

Note: We have to keep in mind that the given equilibrium constant here refers to the concentration of gaseous products. The constant actually tells us when we reach equilibrium do we have more product or more reactant.