At certain temperature- a compound AB2 dissociates as- 2AB2g- 2ABg - B2g with a degree of dissociation - which is very small compared to unity- The expression of Kp in terms of - and total pressure P is -

2AB_2 → 2AB + B_2 1 α α α/2 Total moles = 1 α + α + α/2 = 2 + α/2 P_AB_2 = (1 α)P/(2 + α)/2 = 2 (1 α)P/(2 + α) : P_AB = α P/(2+α)/2 = 2 α ...

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Standard XII

Chemistry

Effective Equilibrium Constant

At certain te...

Question

At certain temperature, a compound $A B_{2}$ dissociates as: $2 A B_{2 (g)} \to 2 A B_{(g)} + B_{2 (g)}$ with a degree of dissociation $α$ , which is very small compared to unity. The expression of $K_{p}$ in terms of $α$ and total pressure P is ?

$\frac{α^{2} P}{2}$

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$\frac{α^{2} P}{3}$

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$\frac{α^{3} P}{3}$

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$\frac{α^{3} P}{2}$

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Solution

The correct option is D
$\frac{α^{3} P}{2}$

$2 A B_{2} \to 2 A B + B_{2}$
$1 - α$ $α$ $\frac{α}{2}$
Total moles $= 1 - α + α + \frac{α}{2} = \frac{2 + α}{2}$
$P_{A B_{2}} = \frac{(1 - α) P}{\frac{(2 + α)}{2}} = \frac{2 (1 - α) P}{(2 + α)} : P_{A B} = \frac{α P}{\frac{(2 + α)}{2}} = \frac{2 α P}{2 + α}; P_{B_{2}} = \frac{\frac{α}{2} P}{\frac{(2 + α)}{2}} = \frac{α P}{2 + α} H e n c e, K_{P} = \frac{P_{A B}^{2} P_{B_{2}}}{P_{A B_{2}}^{2}} = \frac{(\frac{2 α P}{2 + α})^{2}}{[\frac{2 (1 - α) P}{2 + α}]^{2}}$
$= \frac{α^{3} P}{2}$ $(α << 1)$

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At certain temperature, a compound AB2 dissociates as: 2AB2(g)→2AB(g)+B2(g) with a degree of dissociation α, which is very small compared to unity. The expression of Kp in terms of α and total pressure P is ?

The correct option is D α3P2 2AB2→2AB+B2 1−α α α2 Total moles =1−α+α+α2=2+α2 PAB2=(1−α)P(2+α)2=2(1−α)P(2+α):PAB=αP(2+α)2=2αP2+α;PB2=α2P(2+α)2=αP2+αHence,KP=P2ABPB2P2AB2=(2αP2+α)2[2(1−α)P2+α]2 =α3P2 (α<<1)

At certain temperature, a compound $A B_{2}$ dissociates as: $2 A B_{2 (g)} \to 2 A B_{(g)} + B_{2 (g)}$ with a degree of dissociation $α$ , which is very small compared to unity. The expression of $K_{p}$ in terms of $α$ and total pressure P is ?