Question

If the equilibrium constant for the reaction $2AB \to {A_2} + {B_2}$ is $49$, what is the equilibrium constant for $AB \to \dfrac{1}{2}{A_2} + \dfrac{1}{2}{B_2}$
$A.49$
$B.7$
$C.17$
$D.24.5$

Answer 1

this question is an example of the application of the equilibrium constant. The equilibrium constant for the first reaction is already given; use that value to find the equilibrium constant for the second reaction. But before that finding the relationship between the equilibrium constant for the first and the second reaction is important.

Complete answer:
In the above question we are given with two reactions $2AB \to {A_2} + {B_2}$ and $AB \to \dfrac{1}{2}{A_2} + \dfrac{1}{2}{B_2}$. And it is also given that the equilibrium constant for $2AB \to {A_2} + {B_2}$ is $49$. We have to find the equilibrium constant for the reaction $AB \to \dfrac{1}{2}{A_2} + \dfrac{1}{2}{B_2}$
For the first reaction $2AB \to {A_2} + {B_2}$ the equilibrium constant is ${K_1}$. This reaction can also be written as:
${K_1} = \dfrac{{[{A_2}][{B_2}]}}{{{{[AB]}^2}}}$
Let us assume that the reaction for ${K_1}$ is the first equation.
So ${K_1} = \dfrac{{[{A_2}][{B_2}]}}{{{{[AB]}^2}}}\xrightarrow{{}}\left( i \right)$
In the same way for the second reaction $AB \to \dfrac{1}{2}{A_2} + \dfrac{1}{2}{B_2}$ the equilibrium constant is ${K_2}$, and it can also be written as:
${K_2} = \dfrac{{{{[{A_2}]}^{\dfrac{1}{2}}}{{[{B_2}]}^{\dfrac{1}{2}}}}}{{[AB]}}$
Now let us assume this reaction be the second equation.
So ${K_2} = \dfrac{{{{[{A_2}]}^{\dfrac{1}{2}}}{{[{B_2}]}^{\dfrac{1}{2}}}}}{{[AB]}}\xrightarrow{{}}\left( {ii} \right)$
By looking at both the equations $\left( i \right)and\left( {ii} \right)$ it can be seen that the second equation is the square of first equation, hence:
${K_2}^2 = {K_1}$
Now if the second equilibrium constant is the square of the first equilibrium constant. And we know that the value of the first equilibrium constant is $49$. Using this, if we take the square root of the first equilibrium constant we can find the value of the second equilibrium constant.
Thus, ${K_2}^2 = {K_1} = 49$
$\Rightarrow {K_2} = \sqrt {{K_1}}$
$\Rightarrow {K_2} = \sqrt {49}$
$\Rightarrow {K_2} = 7$
Therefore the value of the second equilibrium constant ${K_2}$ for the reaction $AB \to \dfrac{1}{2}{A_2} + \dfrac{1}{2}{B_2}$ is $7$.

Hence the correct option is B.

Note:
The above question is an example of the application of the equilibrium constant. It has many more examples such as it can predict the extent of the reaction; it can also predict the direction of the reaction etc. The value of the equilibrium constant is independent of the initial concentrations of the reacting species. And it changes with the temperature change. it is also independent of the presence of the catalyst.