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Chapter 12: Problem 37

For the reaction $$\mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g)$$ \(K\) is \(1.54 \times 10^{-3}\). When equilibrium is established, the partial pressure of nitrogen is \(0.168 \mathrm{~atm}\), and that of \(\mathrm{NO}\) is \(0.225 \mathrm{~atm}\). The total pressure of the system at equilibrium is \(1.87 \mathrm{~atm}\). What are the equilibrium partial pressures of hydrogen and steam?

Short Answer

Expert verified
Answer: The equilibrium partial pressures of hydrogen (H₂) and steam (H₂O) are approximately 0.699 atm and 0.778 atm, respectively.

Step by step solution

01

Write the equilibrium expression

The equilibrium constant K for the given reaction can be expressed as: $$K = \frac{[NO]^2[H_2]^2}{[N_2][H_2O]^2}$$ where [NO], [H₂], [N₂], and [H₂O] represent the equilibrium partial pressures of the respective gases.
02

Express unknown partial pressures in terms of known values

Since the total pressure at equilibrium is given as 1.87 atm, we can express the partial pressures of H₂ and H₂O in terms of the known partial pressures of N₂ and NO. Let P(H₂) = x and P(H₂O) = y, then: $$P(Total) = P(N_2) + P(H_2O) + P(NO) + P(H_2)$$ $$1.87 = 0.168 + y + 0.225 + x$$ After simplifying, we have: $$x+y=1.477$$
03

Substitute known values and expressions into the equilibrium expression

We now substitute the values and expressions for the partial pressures into the equilibrium expression we wrote in step 1: $$1.54 \times 10^{-3} = \frac{(0.225)^2(x)^2}{(0.168)(y)^2}$$
04

Solve the system of equations

We now have a system of two equations with two unknowns: $$1) \; x + y = 1.477$$ $$2) \; 1.54 \times 10^{-3} = \frac{(0.225)^2(x)^2}{(0.168)(y)^2}$$ First, solve equation (1) for y: $$y = 1.477 - x$$ Now substitute the expression for y into equation (2): $$1.54 \times 10^{-3} = \frac{(0.225)^2(x)^2}{(0.168)(1.477-x)^2}$$
05

Solve for x

After substituting the expression for y, equation (2) becomes a quadratic equation. Solve it for x (the equilibrium partial pressure of H₂): x ≈ 0.699 atm
06

Solve for y

Now substitute the calculated value of x back into the expression for y to solve for the equilibrium partial pressure of H₂O: $$y = 1.477 - 0.699$$ y ≈ 0.778 atm
07

State the answer

The equilibrium partial pressures of hydrogen (H₂) and steam (H₂O) are approximately 0.699 atm and 0.778 atm, respectively.

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Most popular questions from this chapter

WEB At \(1123 \mathrm{~K}\), methane and hydrogen sulfide gases react to form carbon disulfide and hydrogen gases. At equilibrium the concentrations of methane, hydrogen sulfide, carbon disulfide, and hydrogen gas are \(0.00142 M, 6.14 \times 10^{-4} M, 0.00266 M\), and \(0.00943 M\), respectively. (a) Write a balanced equation for the formation of one mole of carbon disulfide gas. (b) Calculate \(K\) for the reaction at \(1123 \mathrm{~K}\).

Consider the following reaction at \(250^{\circ} \mathrm{C}\) : $$\mathrm{A}(s)+2 \mathrm{~B}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{D}(g)$$ (a) Write an equilibrium constant expression for the reaction. Call the equilibrium constant \(K_{1}\). (b) Write an equilibrium constant expression for the formation of one mole of \(\mathrm{B}(\mathrm{g})\) and call the equilibrium constant \(K_{2}\). (c) Relate \(K_{1}\) and \(K_{2}\).

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?

At a certain temperature, the reaction $$\mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightleftharpoons \mathrm{XeF}_{4}(g)$$ gives a \(50.0 \%\) yield of \(\mathrm{XeF}_{4}\), starting with \(\mathrm{Xe}\left(P_{\mathrm{X}_{e}}=0.20 \mathrm{~atm}\right)\) and \(\mathrm{F}_{2}\) \(\left(P_{\mathrm{F}_{2}}=0.40 \mathrm{~atm}\right)\). Calculate \(K\) at this temperature. What must the initial pressure of \(\mathrm{F}_{2}\) be to convert \(75.0 \%\) of the xenon to \(\mathrm{XeF}_{4} ?\)

Given the following reactions and their equilibrium constants, $$\begin{array}{cl}\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) & K=2.4 \times 10^{-9} \\ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) & K=8.8 \times 10^{-13} \end{array}$$ calculate \(K\) for the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$
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