NEET UG 2008 — Chemistry Organic Chemistry

$\begin{array}{cccr} 2 A B_{2(g)} & \rightleftharpoons & 2 A B_{(g)} & + & B_{2(g)} & \\ 2 & & 0 & & 0 & \text { (initially) } \\ 2(1-x) & & 2 x & & x & \text { (at equilibrium) } \end{array}$2AB2(g)​22(1−x)​⇌​2AB(g)​02x​+​B2(g)​0x​ (initially)  (at equilibrium) ​ $\begin{aligned} \text {Amount of moles at equilibrium } & =2(1-x)+2 x+x \\ & =2+x\end{aligned}$Amount of moles at equilibrium ​=2(1−x)+2x+x=2+x​ $\begin{aligned} & K_p= \frac{\left[p_{A B}\right]^2\left[p_{B_2}\right]}{\left[p_{A B_2}\right]^2} \\ & K_p= \frac{\left(\frac{2 x}{2+x} \times P\right)^2 \times\left(\frac{x}{2+x} \times P\right)}{\left(\frac{2(1-x)}{2+x} \times P\right)^2}=\frac{\frac{4 x^3}{2+x} \times P}{4(1-x)^2} \\ & K_p= \frac{4 x^3 \times P}{2} \times \frac{1}{4} \quad(\because 1-x \approx 1 \& 2+x \approx 2) \\ & x=\left(\frac{8 K_p}{4 P}\right)^{1 / 3} \quad \Rightarrow x=\left(\frac{2 K_p}{P}\right)^{1 / 3}\end{aligned}$​Kp​=[pAB2​​]2[pAB​]2[pB2​​]​Kp​=(2+x2(1−x)​×P)2(2+x2x​×P)2×(2+xx​×P)​=4(1−x)22+x4x3​×P​Kp​=24x3×P​×41​(∵1−x≈1&2+x≈2)x=(4P8Kp​​)1/3⇒x=(P2Kp​​)1/3​