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Dalton's Law of Partial Pressures

When NO2 is c...

Question

When $N O_{2}$ is cooled to room temperature, some of it reacts to form a dimer $N_{2} O_{4}$ through the reaction,
$2 N O_{2} (g) \to N_{2} O_{4} (g)$
In the reaction, $15.2 g$ of $N O_{2}$ is placed in a $10.0 L$ flask at high temperature and the flask is cooled to $25^{\circ} C$ . The total pressure is measured to be $0.500 atm$ . Thus, the partial pressure of $N O_{2}$ is found to be:

A

0.31 atm

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B

0.19 atm

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C

0.10 atm

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D

0.40 atm

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Solution

The correct option is B 0.19 atm
Initial moles of $N O_{2} = \frac{15.2}{46} = 0.33$
Molecular weight of $N O_{2} = 14 + 2 \times 16 = 46$
$2 N O_{2} (g) ⇌ N_{2} O_{4} (g)$
After the dimer formation, number of $N$ atom remains unchanged.
Let, $N$ -atom in $N O_{2} = n_{N O_{2}}$
$N$ -atom in $N_{2} O_{4} = 2 n_{N_{2} O_{4}}$ ...(i)
Also by Dalton's law
$P_{N O_{2}} + P_{N_{2} O_{4}} = 0.5 atm$
$\frac{R T}{V} n_{N O_{2}} + \frac{R T}{V} n_{N_{2} O_{4}} = 0.5$
$n_{N O_{2}} + n_{N_{2} O_{4}} = 0.5 \frac{V}{R T}$
$n_{N O_{2}} + n_{N_{2} O_{4}} = \frac{0.5 \times 10}{0.0821 \times 298}$
$n_{N O_{2}} + n_{N_{2} O_{4}} = 0.204$ ...(ii)
Solving Equations (i) and (ii), we get
$n_{N O_{2}} = 0.078 mol n_{N_{2} O_{4}} = 0.126 mol$
Total moles $= 0.078 + 0.126 = 0.204$
Mole fraction of $N O_{2} = \frac{n_{N O_{2}}}{n_{N O_{2}} + n_{N_{2} O_{4}}} = \frac{0.078}{0.204} = 0.38$
$P_{N O_{2}} = χ_{N O_{2}} \times P_{t o t a l} = 0.38 \times 0.5 = 0.191 atm$

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Similar questions

Vapour density of the equilibrium mixture of

N O_{2}

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38.33

. For the equilibrium:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

a. Abnormal molecular weight.

b. Degree of dissociation.

c. Percentage of

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Dalton's Law of Partial Pressures

Standard XII Chemistry

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