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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2025 · 23 January · Evening Shift · Chemistry · Chemical Equilibrium

Consider the reaction X_2 Y(~g)=X_2(~g)+(1)/(2) Y_2(~g) The...

Consider the reaction X2Y( g)=X2( g)+12Y2( g)\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)ofX2Y(g)of \mathrm{X}_2 \mathrm{Y}(\mathrm{g}) with its equilibrium constant Kp is ______ . Assume xx to be very very small.

Options

  1. Ax=2Kpp3x=\sqrt[3]{\frac{2 \mathrm{Kp}}{\mathrm{p}}}
  2. Bx=2Kp2p3x=\sqrt[3]{\frac{2 \mathrm{Kp}^2}{\mathrm{p}}}
  3. Cx=Kpp3x=\sqrt[3]{\frac{\mathrm{Kp}}{\mathrm{p}}}
  4. Dx=Kp2p3x=\sqrt[3]{\frac{K \mathrm{p}}{2 \mathrm{p}}}

Correct answer

B. x=2Kp2p3x=\sqrt[3]{\frac{2 \mathrm{Kp}^2}{\mathrm{p}}}

Step-by-step solution

1 molX2Y(g)→XxX2( g)+12Y2( g)x2∴Px2y=1−x1+x2×pPx2=x1+x2×p\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}} \\ & =\frac{1-\mathrm{x}}{1+\frac{x}{2}} \times \mathrm{p} \\ & P_{x_2}=\frac{x}{1+\frac{x}{2}} \times p\end{aligned}Py2=x/21+x2×p∴Kp=(x1+x2p)(x2(1+x2)p)1/2(1−x1+x2)×p∴Kp=(x1−x)(x2(1+x2))1/2×p1/2\begin{aligned} & P_{y_2}=\frac{x / 2}{1+\frac{x}{2}} \times p \\ \therefore \quad K_p= & \frac{\left(\frac{x}{1+\frac{x}{2}} p\right)\left(\frac{x}{2\left(1+\frac{x}{2}\right)} p\right)^{1 / 2}}{\left(\frac{1-x}{1+\frac{x}{2}}\right) \times p} \\ \therefore \quad & K_p=\left(\frac{x}{1-x}\right)\left(\frac{x}{2\left(1+\frac{x}{2}\right)}\right)^{1 / 2} \times p^{1 / 2} \end{aligned} \therefore \quad \mathrm{x}tobeverysmallto be very small∴Kp=x3/22(1/2)×p1/2∴x3/2=Kp×21/2p1/2x3=Kp2×2px=(Kp2×2p)1/3\begin{aligned} \therefore & K_p=\frac{x^{3 / 2}}{2^{(1 / 2)}} \times p^{1 / 2} \\ \therefore & x^{3 / 2}=\frac{K_p \times 2^{1 / 2}}{p^{1 / 2}} \\ & x^3=\frac{K_p^2 \times 2}{p} \\ & x=\left(\frac{K_p^2 \times 2}{p}\right)^{1 / 3} \end{aligned}$
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