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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

A sample of p...

Question

A sample of pure $N O_{2}$ gas heated to 1000k decomposes : $2 N O_{2} (g) ⇌ 2 N O (g) +)_{2} (g)$ . The equilibrium constant $K_{p}$ is 100 atm . Analysis shows that the partial pressure of $O_{2}$ is 0.25 atm at equilibrium. The partial pressure of $N O_{2}$ at equilibrium is:

0.03

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0.25

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0.025

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0.04

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Solution

The correct option is C 0.025
Pressure $\propto$ moles
Moles of $N O$ = 2 Moles of $O_{2}$
Pressure of $O_{2}$ =0.25
Moles of $O_{2}$ = 0.25
Pressure of $N O$ = 2 X 0.25 = 0.5 atm
$K_{p} = \frac{P O_{2} \times (P_{N O})^{2}}{(P_{N O_{2}})^{2}}$

$100 = \frac{0.25 \times 0.5 \times 0.5}{(P_{N O_{2}})^{2}}$
$P_{N O_{2}} = 0.025$
Option C is correct answer

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A sample of pure NO2 gas heated to 1000k decomposes : 2NO2(g)⇌2NO(g)+)2(g). The equilibrium constant Kp is 100 atm . Analysis shows that the partial pressure of O2 is 0.25 atm at equilibrium. The partial pressure of NO2 at equilibrium is:

The correct option is C 0.025Pressure ∝ molesMoles of NO = 2 Moles of O2Pressure of O2=0.25Moles of O2 = 0.25Pressure of NO = 2 X 0.25 = 0.5 atmKp=PO2×(PNO)2(PNO2)2100=0.25×0.5×0.5(PNO2)2PNO2=0.025Option C is correct answer

A sample of pure $N O_{2}$ gas heated to 1000k decomposes : $2 N O_{2} (g) ⇌ 2 N O (g) +)_{2} (g)$ . The equilibrium constant $K_{p}$ is 100 atm . Analysis shows that the partial pressure of $O_{2}$ is 0.25 atm at equilibrium. The partial pressure of $N O_{2}$ at equilibrium is: