Search any question & find its solution
Question:
Answered & Verified by Expert
The value of equilibrium constant of the reaction $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2$ is 8.0
The equilibrium constant of the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ will be
Options:
The equilibrium constant of the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ will be
Solution:
1612 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{64}$
$\begin{gathered}
\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \\
\mathrm{K}_1=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]} \\
\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\
K_2=\frac{\left[\mathrm{HI}^2\right.}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
\end{gathered}$
From Eqs (i) and (ii)
$\begin{array}{l}
& K_1^2=\frac{1}{K_2} \\
\because & K_1=8.0 \\
\therefore & K_2=\frac{1}{K_1^2}=\frac{1}{8^2}=\frac{1}{64}
\end{array}$
\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \\
\mathrm{K}_1=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]} \\
\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\
K_2=\frac{\left[\mathrm{HI}^2\right.}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
\end{gathered}$
From Eqs (i) and (ii)
$\begin{array}{l}
& K_1^2=\frac{1}{K_2} \\
\because & K_1=8.0 \\
\therefore & K_2=\frac{1}{K_1^2}=\frac{1}{8^2}=\frac{1}{64}
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.