Q8 / 90 · JEE Main 2024 (31 Jan Shift 2) · Chemistry

$A_{(g)} ⇌ B_{(g)} + \frac{C}{2^{(g)}}$ . The correct relationship between $K_{P}, α$ and equilibrium pressure $P$ is

A
$K_{P} = \frac{α^{\frac{1}{2}} P^{\frac{1}{2}}}{{(2 + α)}^{\frac{1}{2}}}$
Correct answer
B
$K_{P} = \frac{α^{\frac{3}{2}} P^{\frac{1}{2}}}{{(2 + α)}^{\frac{1}{2}} (1 - α)}$
C
$K_{P} = \frac{α^{\frac{1}{2}} P^{\frac{3}{2}}}{{(2 + α)}^{\frac{3}{2}}}$
D
$K_{P} = \frac{α^{\frac{1}{2}} P^{\frac{1}{2}}}{{(2 + α)}^{\frac{3}{2}}}$

Solution

$\begin{matrix} A_{(g)} & ⇌ & B_{(g)} & + & \frac{1}{2} C_{(g)} \\ at t = 0 & 1 & 0 & 0 \\ at eq . & 1 - α & α & \frac{α}{2} \end{matrix}$

Total moles at equilibrium: $= 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2}$

$P_{B} = \frac{α}{1 + \frac{α}{2}} p P_{c} = \frac{\frac{α}{2}}{1 + \frac{α}{2}} p P_{A} = \frac{1 - α}{1 + \frac{α}{2}} p$

$K_{p} = \frac{P_{B} \times P_{C}^{\frac{1}{2}}}{P_{A}} = \frac{(\frac{α}{1 + \frac{α}{2}} p) {(\frac{\frac{α}{2}}{1 + \frac{α}{2}} p)}^{\frac{1}{2}}}{(\frac{1 - α}{1 + \frac{α}{2}} p)}$

$K_{p} = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{2^{\frac{1}{2}} (1 - α) {(1 + \frac{α}{2})}^{\frac{1}{2}}}$

$K_{p} = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{(1 - α) {(2 + α)}^{\frac{1}{2}}}$

Hence, the correct answer is option: 'B'.

Ag⇌Bg+C2g. The correct relationship between KP, α and equilibrium pressure P is

Solution

Relevant Questions

$A_{(g)} ⇌ B_{(g)} + \frac{C}{2^{(g)}}$ . The correct relationship between $K_{P}, α$ and equilibrium pressure $P$ is