JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
Consider the reaction \(\mathrm{X}_2 \mathrm{Y}(\mathrm{~g})=\mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g})\)
The equation representing correct relationship between the degree of dissociation (x) of \(\mathrm{X}_2 \mathrm{Y}(\mathrm{g})\) with its equilibrium constant Kp is ______ .
Assume \(x\) to be very very small.
- A \(x=\sqrt[3]{\frac{2 \mathrm{Kp}}{\mathrm{p}}}\)
- B \(x=\sqrt[3]{\frac{2 \mathrm{Kp}^2}{\mathrm{p}}}\)
- C \(x=\sqrt[3]{\frac{\mathrm{Kp}}{\mathrm{p}}}\)
- D \(x=\sqrt[3]{\frac{K \mathrm{p}}{2 \mathrm{p}}}\)
Answer & Solution
Correct Answer
(B) \(x=\sqrt[3]{\frac{2 \mathrm{Kp}^2}{\mathrm{p}}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & 1 \mathrm{~mol} \\ & \mathrm{X}_2 \mathrm{Y}(\mathrm{g}) \rightarrow \underset{\mathrm{x}}{\mathrm{X}} \mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g}) \\ & \frac{\mathrm{x}}{2} \\ & \therefore \quad P_{\mathrm{x}_2 \mathrm{y}} \\ &…
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