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Chapter 12: Problem 40

. 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 \(0.300 \mathrm{~atm} ?\)

Short Answer

Expert verified
Answer: At equilibrium, the partial pressure of N2 is approximately 0.2105 atm, the partial pressure of O2 is approximately 0.2105 atm, and the partial pressure of NO is approximately 0.479 atm.

Step by step solution

01

Write the balanced chemical equation and equilibrium expression.

The balanced chemical equation is given as: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ The equilibrium expression for this reaction, using the equilibrium constant Kp, is: $$K_p = \frac{P_{NO}^2}{P_{N_2}*P_{O_2}}$$ where \(P_{NO}\), \(P_{N_2}\), \(P_{O_2}\) are the partial pressures of NO, N2, and O2, respectively.
02

Set up the ICE table.

An ICE table will help us find the changes in the partial pressures during the reaction. The initial partial pressures of the gases are given as 0.300 atm. | | N2 | O2 | 2NO | |--------|----|----|-----| | Initial|0.300 | 0.300 | 0.300 | | Change | -x | -x | +2x | | Equilibrium|0.300-x|0.300-x|0.300+2x|
03

Substitute equilibrium values into the equilibrium expression and solve for x.

Plugging the equilibrium partial pressures into the equilibrium expression and the given value of Kp (0.0255), we get: $$0.0255 = \frac{(0.300+2x)^2}{(0.300-x)(0.300-x)}$$ This is a quadratic equation in x. We can either solve it by factoring, completing the square, or using the quadratic formula. In this case, solving for x through either method gives: $$x \approx 0.0895$$
04

Calculate the equilibrium partial pressures.

Now, we can substitute x back into our ICE table to find the equilibrium partial pressures of the gases: $$P_{N_2}^{\text{eq}} = 0.300-x = 0.300-0.0895 \approx 0.2105\mathrm{~atm}$$ $$P_{O_2}^{\text{eq}} = 0.300-x = 0.300-0.0895 \approx 0.2105\mathrm{~atm}$$ $$P_{NO}^{\text{eq}} = 0.300+2x = 0.300+2(0.0895) \approx 0.479\mathrm{~atm}$$ So, at equilibrium, the partial pressure of N2 is approximately 0.2105 atm, the partial pressure of O2 is approximately 0.2105 atm, and the partial pressure of NO is approximately 0.479 atm.

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Most popular questions from this chapter

The reaction $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ has an equilibrium constant of \(1.30\) at \(650^{\circ} \mathrm{C}\). Carbon monoxide and steam both have initial partial pressures of \(0.485 \mathrm{~atm}\), while hydrogen and carbon dioxide start with partial pressures of \(0.159\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

The reversible reaction between hydrogen chloride gas and one mole of oxygen gas produces steam and chlorine gas: $$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=0.79$$ Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{HCl}}=P_{\mathrm{O}_{2}}=0.20 \mathrm{~atm}\) (b) \(P_{\mathrm{HCl}}=0.30 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.35 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.15 \mathrm{~atm}\)

Ammonium carbamate solid \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes at \(313 \mathrm{~K}\) into ammonia and carbon dioxide gases. At equilibrium, analysis shows that there are \(0.0451\) atm of \(\mathrm{CO}_{2}, 0.0961\) atm of ammonia, and \(0.159 \mathrm{~g}\) of ammonium carbamate. (a) Write a balanced equation for the decomposition of one mole of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) (b) Calculate \(K\) at \(313 \mathrm{~K}\).

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?

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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