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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

Benzaldehyde, a flavoring agent, is obtained by the dehydrogenation of benzyl alcohol. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(g)+\mathrm{H}_{2}(g)$$ \(K\) for the reaction at \(250^{\circ} \mathrm{C}\) is \(0.56\). If \(1.50 \mathrm{~g}\) of benzyl alcohol is placed in a 2.0-L flask and heated to \(250^{\circ} \mathrm{C}\), (a) what is the partial pressure of the benzaldehyde when equilibrium is established? (b) how many grams of benzyl alcohol remain at equilibrium?

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

Calculate \(K\) for the formation of methyl alcohol at \(100^{\circ} \mathrm{C}\) : $$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ given that at equilibrium, the partial pressures of the gases are \(P_{\mathrm{CO}}=0.814 \mathrm{~atm}, P_{\mathrm{H}_{2}}=0.274 \mathrm{~atm}\), and \(P_{\mathrm{CH}_{3} \mathrm{OH}}=0.0512 \mathrm{~atm} .\)

At \(25^{\circ} \mathrm{C}, K=2.2 \times 10^{-3}\) for the reaction $$\mathrm{ICl}(g) \rightleftharpoons \frac{1}{2} \mathrm{I}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ Calculate \(K\) at \(25^{\circ} \mathrm{C}\) for (a) the decomposition of ICl into one mole of iodine and chlorine. (b) the formation of two moles of \(\operatorname{ICl}(g)\).

Derive the relationship $$K=K_{\mathrm{c}} \times(R T)^{\Delta r_{\mathrm{B}}}$$ where \(K_{\mathrm{c}}\) is the equilibrium constant using molarities and \(\Delta n_{\mathrm{g}}\) is the change in the number of moles of gas in the reaction (see page 326). (Hint: Recall that \(P_{\Lambda}=n_{\Lambda} R T / V\) and \(\left.n_{A} / V=[\mathrm{A}] .\right)\)
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