JEE Main 2025 — Chemistry Physical Chemistry

$\begin{array}{lcc} & \mathrm{AB}_2(\mathrm{g}) & \rightleftharpoons & \mathrm{AB}(\mathrm{g}) & + & \frac{1}{2} \mathrm{~B}_2(\mathrm{g}) \\ t=0 & p_0 & \\ t=t_{\text {eq }} & p_0(1-x) & & p_0 x & & \frac{p_0 x}{2} \end{array}$t=0t=teq ​​AB2​(g)p0​p0​(1−x)​⇌​AB(g)p0​x​+​21​ B2​(g)2p0​x​​ $\mathrm{p}=\mathrm{p}_0 \mathrm{x}-\mathrm{p}_0 \mathrm{x}+\mathrm{p}_0 \mathrm{x}+\frac{\mathrm{p}_0 \mathrm{x}}{2}$p=p0​x−p0​x+p0​x+2p0​x​ $\begin{gathered}p=p_0\left(1+\frac{x}{2}\right) \\ p_0=\frac{p}{\left(1+\frac{x}{2}\right)} \\ K_p=\frac{\left(p_{A B}\right)\left(p_{B_2}\right)^{1 / 2}}{\left(p_{\mathrm{AB}_2}\right)} \\ K_p=\frac{\left(p_0 x\right)\left(\frac{p_0 x}{2}\right)^{1 / 2}}{p_0(1-x)}\end{gathered}$p=p0​(1+2x​)p0​=(1+2x​)p​Kp​=(pAB2​​)(pAB​)(pB2​​)1/2​Kp​=p0​(1−x)(p0​x)(2p0​x​)1/2​​ $\mathrm{K}_{\mathrm{p}}=\frac{\frac{\mathrm{px}}{\left(1+\frac{x}{2}\right)}\left(\frac{\mathrm{p}}{1+\frac{x}{2}} \times \frac{\mathrm{x}}{2}\right)^{1 / 2}}{\frac{\mathrm{p}(1-\mathrm{x})}{\left(1+\frac{x}{2}\right)}}$Kp​=(1+2x​)p(1−x)​(1+2x​)px​(1+2x​p​×2x​)1/2​ Since $x \ll 1$x≪1 $\begin{aligned}
& K_p=\frac{\mathrm{p}^{1 / 2} \mathrm{x}^{3 / 2}}{2^{1 / 2}} \
& \mathrm{x}=\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}}
\end{aligned}$