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Dalton's Law of Partial Pressures

At 1000K, a...

Question

At $1000$ K, a sample of pure $N O_{2}$ gas decomposes as:
$2 N O_{2} (g) ⇌ 2 N O (g) + O_{2} (g)$
The equilibrium constant $K_{p}$ is $156.25$ atm. Analysis shows that the partial pressure $0.25$ atm at equilibrium. The partial pressure of $N O_{2}$ at equilibrium is:

0.01

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0.02

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0.04

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None of the above

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Solution

The correct option is B $0.02$
${2 N O}_{2} (g) ⇌ 2 N O (g) + O_{2} (g)$

$2 : 1$ ratio between $N O$ and $O_{2}$ .

Given, the partial pressure of oxygen is $0.25$ atm.

the partial pressure of $N O$ is $= 2 (0.25) = 0.5$ atm

$K_{P} = \frac{{(P_{N O})}^{2} (P_{O_{2}})}{{(P_{{N O}_{2}})}^{2}} P_{{N O}_{2}} = P A s s u m e$

$K_{P} = 156.25$ atm

$156.25 = \frac{{(0.5)}^{2} (0.25)}{P^{2}}$

$P^{2} = \frac{0.0625}{156.25} = 4 \times 10^{- 4}$

$P = 2 \times 10^{- 2} = 0.02$ atm.

Option - B is correct.

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