Question

For the reaction, ${{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{I}}_{\text{2}}}\left( {\text{g}} \right) \rightleftharpoons 2{\text{HI}}\left( {\text{g}} \right)$ , the equilibrium constant ${{\text{K}}_{\text{p}}}$ changes with:
A.Total pressure
B.Catalyst
C.The amounts of \[{{\text{H}}_{\text{2}}}\] and \[{{\text{I}}_{\text{2}}}\] present
D.Temperature

Answer 1

Hint: The equilibrium constant in terms of partial pressure ${{\text{K}}_{\text{p}}}$ is related to the equilibrium constant in terms of concentration ${{\text{K}}_{\text{c}}}$ as follows:
${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}} \times {\text{R}}{{\text{T}}^{\Delta {\text{n}}}}$
Here, $\Delta {\text{n}}$ is the change in the number of moles of the gaseous products and the gaseous reactants.
A catalyst is a substance which helps to increase the rate of a chemical reaction without itself undergoing any change. It does not disturb the equilibrium of the reaction.

Complete step by step answer:
For the reaction given in the question ${{\text{H}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{I}}_{\text{2}}}\left( {\text{g}} \right) \rightleftharpoons 2{\text{HI}}\left( {\text{g}} \right)$ , we can see that there are two moles of product hydrogen iodide which is in gaseous state and there are one mole each of the reactants hydrogen and iodine, both of which are also in gaseous state. Thus, the total mole of gaseous reactants is also 2. Hence, the change in the number of moles of the gaseous products and the gaseous reactants is zero. Hence, $\Delta {\text{n}}$ is equal to 0.
The term RT is raised to the power $\Delta {\text{n}}$ . Since anything raised to the power zero is equal to 1, thus, RT raised to the power zero is also equal to 1. Hence, equilibrium constant in terms of partial pressure ${{\text{K}}_{\text{p}}}$ is equal to the equilibrium constant in terms of concentration ${{\text{K}}_{\text{c}}}$ . Thus,
${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}}$
Now, the relation of ${{\text{K}}_{\text{c}}}$ with temperature is as follows:
$\log \dfrac{{{{\text{K}}_{\text{2}}}}}{{{{\text{K}}_{\text{1}}}}} = \dfrac{{\Delta {\text{H}}}}{{{\text{2}}{\text{.303R}}}}\left[ {\dfrac{{\text{1}}}{{{{\text{T}}_{\text{1}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{T}}_{\text{2}}}}}} \right]$
Here, ${{\text{K}}_{\text{2}}} = $ equilibrium constant at temperature ${{\text{T}}_{\text{2}}}$
${{\text{K}}_1} = $ equilibrium constant at temperature ${{\text{T}}_1}$
R is the gas constant and $\Delta {\text{H}}$ is enthalpy change in the temperature range ${{\text{T}}_1}$ to ${{\text{T}}_{\text{2}}}$ .
Hence, we see that the equilibrium constant has a definite value for every reaction at a given temperature. But, it varies with change in temperature. Hence, option D is correct.
Also, we see that the equilibrium constant is independent of the total pressure and also of the presence of a catalyst. It is also independent of the original concentrations of the reactants. Hence, A, B, and C are incorrect.

Hence, option D is correct.

Note:
One other characteristic of the equilibrium constant is that the equilibrium constant for the forward reaction is equal to the inverse of the equilibrium constant for the backward reaction.
The value of ${{\text{K}}_{\text{c}}}$ or ${{\text{K}}_{\text{p}}}$ tells the extent to which the forward or the backward reaction has taken place. Higher value of ${{\text{K}}_{\text{c}}}$ or ${{\text{K}}_{\text{p}}}$ means that the reaction has proceeded to a greater extent in the forward direction.