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Chapter 12: Problem 53

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

Short Answer

Expert verified
Answer: The new equilibrium partial pressures of the gases are: - $$P_{N_{2}(new)} = 1.0667\mathrm{~atm}$$ - $$P_{O_{2}(new)} = 1.0667\mathrm{~atm}$$ - $$P_{NO(new)} = 0.2886\mathrm{~atm}$$

Step by step solution

01

Write the expression for the equilibrium constant K_p

For the given reaction: $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g})$$ The equilibrium constant in terms of partial pressures, \(K_p\), can be expressed as: $$K_p = \frac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}$$
02

Calculate K_p using the given partial pressures

Using the given partial pressures: $$P_{\mathrm{N}_{2}}= 1.2 \mathrm{~atm}$$ $$P_{\mathrm{O}_{2}}= 0.80 \mathrm{~atm}$$ $$P_{\mathrm{NO}}= 0.022 \mathrm{~atm}$$ We can now calculate \(K_p\): $$K_p = \frac{(0.022)^2}{(1.2)(0.8)}= \frac{0.000484}{0.96} = 5 \times 10^{-4}$$
03

Calculate the change in oxygen pressure and the system's new reaction quotient Q_p

The problem states that more oxygen is added to the system, increasing its partial pressure to \(1.2 \mathrm{~atm}\). We're supposed to find the new equilibrium partial pressures of all gases. First, calculate the change in \(P_{O_2}\): $$\Delta P_{O_2} = 1.2 \mathrm{~atm} - 0.8 \mathrm{~atm} = 0.4 \mathrm{~atm}$$ Let's assume x amount of \(O_2\) is consumed upon reaching the new equilibrium, the new partial pressures will be: $$P_{N_2} = 1.2 - x \mathrm{~atm}$$ $$P_{O_2} = 1.2 - x \mathrm{~atm}$$ $$P_{NO} = 0.022 + 2x \mathrm{~atm}$$ Now, let's write the expression for the new reaction quotient, \(Q_p\): $$Q_p = \frac{(0.022 + 2x)^2}{(1.2 - x)(1.2 - x)}$$
04

Set Q_p = K_p and solve for x

Since the system is at equilibrium, \(Q_p = K_p\). Therefore: $$\frac{(0.022 + 2x)^2}{(1.2 - x)^2} = 5 \times 10^{-4}$$ We'll solve for x: $$x = \frac{|\Delta P_{O_2}|}{3} = \frac{0.4 \mathrm{~atm}}{3} = 0.1333 \mathrm{~atm}$$
05

Calculate the new partial pressures at equilibrium

Now that we have the x value, we can find the new partial pressures at equilibrium: $$P_{N_{2}(new)} = 1.2\mathrm{~atm} - 0.1333\mathrm{~atm} = 1.0667\mathrm{~atm}$$ $$P_{O_{2}(new)} = 1.2\mathrm{~atm} - 0.1333\mathrm{~atm} = 1.0667\mathrm{~atm}$$ $$P_{NO(new)} = 0.022\mathrm{~atm} + 2(0.1333)\mathrm{~atm} = 0.2886\mathrm{~atm}$$ The new equilibrium partial pressures of all gases are: $$P_{N_{2}(new)} = 1.0667\mathrm{~atm}$$ $$P_{O_{2}(new)} = 1.0667\mathrm{~atm}$$ $$P_{NO(new)} = 0.2886\mathrm{~atm}$$

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Most popular questions from this chapter

. Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ At a certain temperature, the equilibrium constant for the reaction is \(0.0255\). What are the partial pressures of all gases at equilibrium if the initial partial pressure of the gases (both reactants and products) is \(0.300 \mathrm{~atm} ?\)

Iodine chloride decomposes at high temperatures to iodine and chlorine gases. $$2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g)$$ Equilibrium is established at a certain temperature when the partial pressures of ICI, \(\mathrm{I}_{2}\), and \(\mathrm{Cl}_{2}\) are (in atm) \(0.43,0.16\), and \(0.27\), respectively. (a) Calculate \(K\). (b) If enough iodine condenses to decrease its partial pressure to \(0.10 \mathrm{~atm}\), in which direction will the reaction proceed? What is the partial pressure of iodine when equilibrium is re-established?

For the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) at a certain temperature is \(0.50\). Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{O}_{2}}=P_{\mathrm{NO}}=P_{\mathrm{NO}_{2}}=0.10 \mathrm{~atm}\) (b) \(P_{\mathrm{NO}_{2}}=0.0848 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.0116 \mathrm{~atm}\) (c) \(P_{\mathrm{NO}_{2}}=0.20 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.010 \mathrm{~atm}, P_{\mathrm{NO}}=0.040 \mathrm{~atm}\)

Given the following descriptions of reversible reactions, write a balanced net ionic equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Liquid acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) is in equilibrium with its vapor. (b) Hydrogen gas reduces nitrogen dioxide gas to form ammonia and steam. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

The system $$3 \mathrm{Z}(g)+\mathrm{Q}(g) \rightleftharpoons 2 \mathrm{R}(g)$$ is at equilibrium when the partial pressure of \(\mathrm{Q}\) is \(0.44 \mathrm{~atm} .\) Sufficient \(\mathrm{R}\) is added to increase the partial pressure of Q temporarily to \(1.5 \mathrm{~atm} .\) When equilibrium is reestablished, the partial pressure of \(Q\) could be which of the following? (a) \(1.5 \mathrm{~atm}\) (b) \(1.2 \mathrm{~atm}\) (c) \(0.80\) atm (d) \(0.44 \mathrm{~atm}\) (e) \(0.40 \mathrm{~atm}\)
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