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Chapter 15: Problem 65

Consider the reaction in a closed container: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Initially, 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(\alpha\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(\alpha\) and \(P,\) the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium caused by an increase in \(P ?\) Does your prediction agree with Le Châtelier's principle?

Short Answer

Expert verified
The derived expression for \(K_{P}\) is \(K_{P} = P{4\alpha^{2} \over (1-\alpha)(1+\alpha)}\). As per this expression, an increase in total pressure will shift the equilibrium towards the right (towards \(NO_{2}\)), which is in agreement with Le Châtelier's principle.

Step by step solution

01

Initial conditions

Initially, there is 1 mole of \(N_{2}O_{4}\) gas and none of the \(NO_{2}\) gas in the system. Since the reaction hasn't started, no \(N_{2}O_{4}\) has dissociated.
02

Conditions at equilibrium

At equilibrium, \(\alpha\) mole of \(N_{2}O_{4}\) has dissociated into \(NO_{2}\). Since each mole of \(N_{2}O_{4}\) produces 2 moles of \(NO_{2}\), there will be \(2\alpha\) moles of \(NO_{2}\) gas formed and \(1-\alpha\) mole of \(N_{2}O_{4}\) remaining.
03

Expressing partial pressures

The partial pressure of a gas in a mixture is given by the mole fraction of the gas multiplied by the total pressure of the system. So, we can express the partial pressures of \(N_{2}O_{4}\) and \(NO_{2}\) as follows: \[ P(N_{2}O_{4}) = P{1-\alpha \over 1 + \alpha} \] \[ P(NO_{2}) = P{2\alpha \over 1 + \alpha} \] Here \(P\) is the total pressure of the system.
04

Deriving the expression for \(K_{P}\)

The equilibrium constant \(K_P\) is given by the ratio of the products of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. For this reaction: \[ K_{P} = {(P(NO_{2}))^2 \over P(N_{2}O_{4})} \] Substituting the expressions for partial pressures from step 3 into this equation, we get \[ K_{P} = P{(2\alpha)^{2} \over (1 - \alpha)}{1 \over (1 + \alpha)} \] which simplifies to \(K_{P} = P{4\alpha^{2} \over (1-\alpha)(1+\alpha)} \)
05

Predicting the shift in equilibrium with change in total pressure

From the derived expression, it can be seen that \(K_{P}\) is directly proportional to the total pressure \(P\). Hence, increasing \(P\) will increase \(K_{P}\), and to maintain the equilibrium (because \(K_{P}\) is a constant at a given temperature), the reaction will shift in the direction where the total gas moles (and hence pressure) are increased. In this case, it is towards the right (towards \(NO_{2}\)).
06

Comparison with Le Châtelier's principle

Le Châtelier's principle states that if a system at equilibrium is subjected to a change, the system will adjust itself in such a way as to counteract that change. In this case, an increase in pressure would shift the equilibrium towards the side with more moles of gas, which is consistent with our prediction from the derived expression. This confirms that the derived expression and Le Châtelier's principle are in agreement.

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Most popular questions from this chapter

Consider this reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{P}\) for the reaction is \(1.0 \times\) \(10^{-15}\) at \(25^{\circ} \mathrm{C}\) and 0.050 at \(2200^{\circ} \mathrm{C}\). Is the formation of nitric oxide endothermic or exothermic? Explain your answer.

Define homogeneous equilibrium and heterogeneous equilibrium. Give two examples of each.

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) $$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}, 1.35 \times\) \(10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask at equilibrium. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

Consider the following reaction at \(1600^{\circ} \mathrm{C}\) : $$ \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) $$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 - \(\mathrm{L}\) flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.
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