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Dalton's Law of Partial Pressures

A sample of p...

Question

A sample of pure $N O_{2}$ gas healed lo 1000 K decomposes.
$N O_{2} (g) ⇋ 2 N O (g) + O_{2} (g)$
The equillibrium constant $K_{P}$ is 100atm. Analysis shows that the partial pressure of $O_{2}$ is 0.25 atm at equillibrium. The partial pressure of $N O_{2}$ at equillibrium is:

0.03

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0.25

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0.0245

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0.04

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Solution

The correct option is C 0.03
$2 N O_{2} (g) ⇋ 2 N O (g) + O_{2} (g)$
Analysis shows that the partial pressure of $O_{2}$ is 0.25 atm at equilibrium. The partial pressure of $N O$ at equilibrium $2 \times$ the partial pressure of $O_{2}$ $2 \times 0.25 = 0.50$ atm.
The equillibrium constant $K_{P}$ is 100 atm.
$K_{p} = \frac{P_{N O}^{2} \times P_{O 2}}{P_{N O 2}^{2}}$
$100 = \frac{(0.50)^{2} \times 0.25}{P_{N O 2}^{2}}$
$P_{N O 2}^{2} = 0.000625$
$P_{N O 2} = 0.025 ≃ 0.03$
The partial pressure of $N O_{2}$ at equilibrium is 0.03 atm.

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A sample of pure NO2 gas healed lo 1000 K decomposes.NO2(g)⇋2NO(g)+O2(g)The equillibrium constant KP is 100atm. Analysis shows that the partial pressure of O2 is 0.25 atm at equillibrium. The partial pressure of NO2 at equillibrium is:

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$N O_{2} (g) ⇋ 2 N O (g) + O_{2} (g)$
The equillibrium constant $K_{P}$ is 100atm. Analysis shows that the partial pressure of $O_{2}$ is 0.25 atm at equillibrium. The partial pressure of $N O_{2}$ at equillibrium is: