Question

For a chemical reaction of the type $A \rightleftharpoons B$, K=2 and $B
\rightleftharpoons C$, K=0.01
Equilibrium constant for the reaction $2C \rightleftharpoons 2{\mathbf{A}}$ is
(a) 25
(b) 50
(c) 2500
(d) $4 \times {10^{ - 4}}$

Answer 1

Hint: We know that the equilibrium constant is the ratio of the product of molar concentrations of the products to the product of molar concentration of the reactants, each raised to the power equal to its stoichiometric coefficient at constant temperature. Equilibrium constant in terms of concentration is represented as ${K_c}$.

Complete Step by step answer:
Here we can see that the given reactions all are first order reactions, also the order of reaction doesn’t matter here. We just have to relate somehow the equilibrium constant of the given reactions.
Let us take a look at the given reaction
$A \rightleftharpoons {\text{B}}$
From the definition of the equilibrium constant we can say that
${K_1} = \dfrac{{\left[ B \right]}}{{\left[ A \right]}}$ ------(1)
Let ${K_1}$ be the equilibrium constant of first reaction, also given that
${K_1} = 2$
Now for the second reaction,
$B \rightleftharpoons {\text{C}}$
${K_2} = \dfrac{{\left[ C \right]}}{{\left[ B \right]}}$ --------(2)
Where ${K_2}$ is the equilibrium constant of second reaction, also given that
${K_2} = 0.01$
We have to find the equilibrium constant of the reaction $2C \rightleftharpoons 2A$
The equilibrium constant of this reaction will be
${K_3} = \dfrac{{{{\left[ A \right]}^2}}}{{{{\left[ C \right]}^2}}}$
Where ${K_3}$ be the equilibrium constant for the third reaction.
Now can compare the equations (1) and (2)
For finding the equilibrium constant of final reaction, we do not need the concentration of B so in reaction (1) and (2) we can equate them so that we get a relation between A and C.
Rewriting the equation (1) and (2)
$2 = \dfrac{{\left[ B \right]}}{{\left[ A \right]}}\;\,\, \Rightarrow \left[ B \right] = 2\left[ A \right]$ ------------(3)
$0.01 = \dfrac{{\left[ C \right]}}{{\left[ B \right]}}\,\, \Rightarrow \left[ B \right] = \dfrac{{\left[ C \right]}}{{0.01}}$ --------------(4)
Now, we can equate (3) and (4), we will get
$2\left[ A \right] = \dfrac{{\left[ C \right]}}{{0.01}}$
On cross multiplying, we get
$\dfrac{{\left[ A \right]}}{{\left[ C \right]}} = \dfrac{1}{{0.02}}$
$ \Rightarrow \dfrac{{\left[ A \right]}}{{\left[ C \right]}} = \dfrac{{100}}{2} = 50$
Now, squaring on both sides we get
$\dfrac{{{{\left[ A \right]}^2}}}{{{{\left[ C \right]}^2}}} = {50^2} = 2500$
$\Rightarrow {K_3} = 2500$
Hence we got the equilibrium constant as 2500

Hence, Option (c) is correct.

Note: Often what happens is that we forget to balance our equation, which results in wrong powers raised to the reactant or products. Here even though without the 2 moles of reactants and products they were balanced but since it is given in the question we have to take 2 as the powers for its concentration.