If represents the equilibrium pressure and a be the degree of dissociation, th en match the following Column - I Column - II (Equilibrium type) (K P relations) (P) K P is independent of P (1) PCl 5

Question

If represents the equilibrium pressure and a be the degree of dissociation, th en match the following

Column - I Column - II

(Equilibrium type) (K_P relations)

(P) $K_{P}$ is independent of P (1) $PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)$

(Q) $K_{P} = \frac{27 α ^{4} P ^{2}}{16 ( 1 - α ^{2} ) ^{2}}$ (2) $N_{2} O_{4} (g) ⇌ 2 NO_{2} (g)$

(R) $K_{P} = \frac{α ^{2} P}{1 - α ^{2}}$ (3) $2 ClF_{3} (g) ⇌ Cl_{2} (g) + 3 F_{2} (g)$

(S) $K_{P} = \frac{4 α ^{2} P}{1 - α ^{2}}$ (4) $NO (g) ⇌ 1/2 N_{2} (g) + 1/2 O_{2} (g)$

TestHub · Accepted Answer

Explaining the Question :- Match the column

Concept:- This question is based on Degree of dissociation

Solution :- $\displaylines NO_{(g)} ⇌ 21 N_{2 (g)} + \frac{1}{2} O_{2 (g)} \frac{α}{2} (∴ Kp = Kc ∵ Δ ng = 0)$

1 - -

$1 - α$ $\frac{α}{2}$ $\frac{α}{2}$

$n_{T} = 1 - α + \frac{α}{2} + \frac{α}{2} = 1$

$K_{p} = \frac{( \frac{α}{2} P _{T} ) ^{\frac{1}{2}} ( \frac{α}{2} P _{T} ) ^{\frac{1}{2}}}{( 1 - α ) P _{T}} = \frac{α}{2 ( 1 - α )} (P \to 4)$

(Q) $2 ClF_{3} ⇌ Cl_{2 (g)} + 3 F_{2}$

$\displaylines 1 1 - α - \frac{α}{2} - \frac{3 α}{2}$

$n_{T} = 1 - α + \frac{α}{2} + \frac{3}{2} α$

$= 1 + α$

$K_{p} = \frac{( \frac{α}{2 ( 1 + α )} P ) ( \frac{3 α}{2 ( 1 + α )} P ) ^{3}}{( \frac{1 - α}{1 + α} P ) ^{2}} = \frac{27 α ^{4} P ^{2}}{16} (Q \to 3)$

(R) $PCl_{5 (g)} ⇌ PCl_{3 (g)} + Cl_{2 (g)}$

1 - -

$1 - α$ $α$ $α$ $n_{T} = 1 + α$

$K_{p} = \frac{( \frac{α}{1 + α} P ) ( \frac{α}{1 + α} P )}{( \frac{1 - α}{1 + α} P )} = \frac{α ^{4} P}{1 - α ^{2}} (R \to 1)$

(S) $N_{2} O_{4 (g)} ⇌ 2 NO_{2 (g)}$

$1$ -

$1 - α$ $2 α$ $n_{T} = 1 + α$

$K_{p} = \frac{( \frac{α}{1 + α} P ) ^{2}}{( \frac{1 - α}{1 + α} P )} = \frac{4 α ^{2} P}{1 - α ^{2}}$

Final Answer :- D

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