Question

At 700K, equilibrium constant for the reaction:${{H}_{2}}(g)+{{I}_{2}}(g)\rightleftarrows 2HI$ is 64. If 0.5 mol${{L}^{-1}}$ of HI (g) is present at 700K, what are the concentration of ${{H}_{2}}(g)$ and ${{I}_{2}}(g)$ assuming that we initially started with HI (g) and allowed it to reach equilibrium at 700K ?
A. 0.0625M
B. 0.034M
C. 0.136M
D. 0.068M

Answer 1

Hint: As we know that the solubility product of an electrolyte at a particular temperature is equal to the product of the molar concentration of its ions in the saturated solution, and each concentration raised to power equal to the number of ions produced on dissociation.

Complete Solution :
 - We are being provided in the question with the equilibrium constant ${{K}_{C}}$ value for the reaction: ${{H}_{2}}(g)+{{I}_{2}}(g)\rightleftarrows 2HI$ as 54.8.
- Hence, at equilibrium, we can write the equilibrium constant ${{{K}'}_{C}}$ for the reaction :
\[2HI\left( g \right)\rightleftarrows {{H}_{2}}\left( g \right)+{{I}_{2}}\left( g \right)\]
 as 1/54.8

- We are also being provided with the value of equilibrium concentration of HI as 0.5 M.
- Now, let us consider the concentrations of iodine and hydrogen at equilibrium be x mol${{L}^{-1}}$.
Therefore,
$[{{H}_{2}}]=[{{I}_{2}}]=x\text{ }mol{{L}^{-1}}$

- We can write the equation for equilibrium constant as:
$\begin{align}
 & \dfrac{[{{H}_{2}}][{{I}_{2}}]}{[HI]}={{{{K}'}}_{C}} \\
 & \Rightarrow \frac{x\times x}{{{\left( 0.5 \right)}^{2}}}=\dfrac{1}{54.8} \\
 & \Rightarrow {{x}^{2}} = \dfrac{0.25}{54.8} \\
 & \Rightarrow {{x}^{2}} = 0.068mol{{L}^{-1}} \\
\end{align}$- Hence, we can conclude that the correct option is (d), that is the concentration of ${{H}_{2}}(g)$ and ${{I}_{2}}(g)$ is 0.068M. So, the correct answer is “Option D”.

Note: - It is found that when a strong electrolyte is added (having a common ion) in a solution of weak electrolyte then the ionization of that weak electrolyte is suppressed.
- It is found that the change of temperature is the only one thing that changes the value of equilibrium constant.