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Chapter 15: Problem 54

Pure NOCl gas was heated at \(240^{\circ} \mathrm{C}\) in a 1.00 - \(\mathrm{L}\) container. At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm. $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) in the system. (b) Calculate the equilibrium constant \(K_{P}\).

Short Answer

Expert verified
The partial pressures of NO and Cl2 are 0.24 atm and 0.12 atm respectively. The value of the equilibrium constant \(K_{P}\) is 0.75.

Step by step solution

01

Determine Partial Pressures

At equilibrium, we know that the pressure of NOCl is 0.64 atm and the total pressure is 1 atm. Therefore, the partial pressures of NO and Cl2 combined is \(1 - 0.64 = 0.36 atm\). Since the reaction produces 2 moles of NO and 1 mole of Cl2, we would expect twice as much NO as Cl2. Therefore, the partial pressure of NO is \(2/3 * 0.36 = 0.24 atm\) and the partial pressure of Cl2 is \(1/3 * 0.36 = 0.12 atm\).
02

Determine the Pressure New Mixture

Use the equation for the equilibrium constant for pressure (Kp), which is equal to the products of the partial pressures of the products raised to their stoichiometric coefficients, divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients. So, according to the balanced chemical equation, we see that Kp should be equal to (P(NO)^2 * P(Cl2)) / (P(NOCl)^2).
03

Calculate the Equilibrium Constant

Substitute the determined pressures into the \(K_{P}\) equation: \(K_{P} = (0.24^{2} * 0.12) / 0.64^{2}= 0.75\).

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Most popular questions from this chapter

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures of \(0.177 \mathrm{~atm}\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

Consider the reaction $$ \begin{aligned} 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & \\ \Delta H^{\circ}=&-198.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to \((\mathrm{a})\) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2},\) (d) add a catalyst, (e) add helium at constant volume.

What effect does an increase in pressure have on each of these systems at equilibrium? (a) \(\mathrm{A}(s) \rightleftharpoons 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftharpoons \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftharpoons \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\) The temperature is kept constant. In each case, the reacting mixture is in a cylinder fitted with a movable piston.

The dissociation of molecular iodine into iodine atoms is represented as $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K}\), the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(I_{2}\) in a 2.30 - \(L\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.
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