Questions

For following equilibrium: $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ ; $K_{P} = \frac{P_{{NO}_{2}}^{2}}{P_{N_{2} O_{4}}}$ ; Graph between $P_{total} V s$ $α$ is given. Where $P_{t o t a l}$ = Total pressure at equilibrium a ® degree of dissociation (a << 1)

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value of

K_{P}

36 a t m .

value of

P_{2}

56.25 a t m .

If total pressure is increased at equilibrium then reaction will move in backward direction.

On increasing temperature degree of dissociation of

N 2 O 4 (g)

will increase.

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detailed solution

Correct option is A

$\underset{a_{0} - a_{0} α}{N_{2} O_{4} (g)} ⇌ \underset{2 a_{0} α}{2 {NO}_{2} (g)} K_{P} = \frac{4 α^{2}}{1 - α^{2}} \cdot P_{t} \approx 4 α^{2} \cdot P_{t} if α < < 1$

$(1) K_{P} = 4 \times (0.3)^{2} \times 100 = 36 atm K_{P} = 4 α_{2}^{2} P_{2} 4 \times (0.3)^{2} \times 100 = 4 (0.4)^{2} P_{2}$

$(2) P_{2} = \frac{900}{16} = 56.25 atm$

$(3) Q_{P} = 4 α^{2} . P_{t}$ so if $P_{t}$ increases then $α$ decreases to reset $Q_{P} = K_{P} .$

$(4)$ As reaction is endothermic in forward direction so on increasing temperature, it favour endothermic step i.e. $α$ increases.

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For following equilibrium: N2O4( g)⇌2NO2( g) ; KP=PNO22PN2O4; Graph between Ptotal Vs α is given. Where Ptotal = Total pressure at equilibrium a ® degree of dissociation (a << 1)