Question

The reaction below was carried out at a pressure of $10.00 \times 10^{4} Pa$ and at constant temperature.

$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$ .

The partial pressures of nitrogen and oxygen are both $4.85 \times 10^{4} Pa$ . Calculate the partial pressure of the nitrogen $(II)$ oxide, $NO (g)$ , at equilibrium.

Accepted Answer

The pressure exerted by an individual gas in a mixture is known as its partial pressure. Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases: $P_{Total} = p_{1} + p_{2} + p_{3}$ .

Given,

The pressure of the reaction, $P_{Total} = 10.00 \times 10^{4} Pa$ .

The partial pressures of nitrogen, $p_{N_{2}} = 4.85 \times 10^{4} Pa$ .

The partial pressures of oxygen, $p_{O_{2}} = 4.85 \times 10^{4} Pa$ .

The partial pressures of nitrogen $(II)$ oxide, $p_{NO} = ?$

$P_{Total} = p_{N_{2}} + p_{O_{2}} + p_{NO} . 10.00 \times 10^{4} = 4.85 \times 10^{4} + 4.85 \times 10^{4} + p_{NO} . p_{NO} = 10.00 \times 10^{4} - (4.85 \times 10^{4} + 4.85 \times 10^{4}) . p_{NO} = 10.00 \times 10^{4} - 9.7 \times 10^{4} = 0.3 \times 10^{4} Pa .$

Lawrie Ryan and Roger Norris Solutions for Chapter: Equilibria, Exercise 9: Questions

Lawrie Ryan Chemistry Solutions for Exercise - Lawrie Ryan and Roger Norris Solutions for Chapter: Equilibria, Exercise 9: Questions

Questions from Lawrie Ryan and Roger Norris Solutions for Chapter: Equilibria, Exercise 9: Questions with Hints & Solutions

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