A_ (g) B_ (g) + C 2 (~g) The correct relationship between K_P, an… - Vidyadip

MCQ

$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g}) \quad$ The correct relationship between $K_P, \alpha$ and equilibrium pressure $P$ is

A
$K_P=\frac{\alpha^{1 /_2} P^{1 / 2}}{(2+\alpha)^{1 /_2}}$
✓
$K_P=\frac{\alpha^{3 /_2} P^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}$
C
$K_P=\frac{\alpha^{1 /_2} P^{3 / 2}}{(2+\alpha)^{3 /_2}}$
D
$K_P=\frac{\alpha^{1 /_2} P^{1 / 2}}{(2+\alpha)^{3 /_2}}$

✓

Answer

Correct option: B.

$K_P=\frac{\alpha^{3 /_2} P^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}$

b
$A_{(g)} \rightleftharpoons B_{(g)}+\frac{C}{2}(g)$

$t=t_{e q}(1-\alpha) \quad \alpha \quad \frac{\alpha}{2}$

$\mathrm{P}_{\mathrm{B}}=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} . \mathrm{P}, \mathrm{P}_{\mathrm{A}}=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{C}}=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}$

$K_P=\frac{P_B \cdot P_C^{\frac{1}{2}}}{P_A}$

$=\frac{(\alpha)^{\frac{3}{2}}(P)^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}}$

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