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Chapter 12: Problem 35

At a certain temperature, \(K\) is \(1.3 \times 10^{5}\) for the reaction $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ What is the equilibrium pressure of hydrogen sulfide if those of hydrogen and sulfur gases are \(0.103\) atm and \(0.417\) atm, respectively?

Short Answer

Expert verified
Question: Given the reaction H2(g) + S2(g) ⇌ 2H2S(g) with an equilibrium constant K = 1.3 x 10^5, find the equilibrium pressure of H2S if the equilibrium pressures of H2 and S2 are 0.103 atm and 0.417 atm, respectively. Answer: The equilibrium pressure of hydrogen sulfide (H2S) is approximately 0.449 atm.

Step by step solution

01

Write down the expression for the equilibrium constant (K)

According to the given balanced chemical equation, the expression for the equilibrium constant (K) can be written as follows: $$K = \frac{P_{H_2S}^2}{P_{H_2}^2 \times P_{S_2}}$$ Here, \(P_{H_2S}\) is the equilibrium pressure of hydrogen sulfide, \(P_{H_2}\) is the equilibrium pressure of hydrogen, and \(P_{S_2}\) is the equilibrium pressure of sulfur.
02

Insert the given values into the expression for K

We have K = \(1.3 \times 10^{5}\), \(P_{H_2} = 0.103\) atm, and \(P_{S_2} = 0.417\) atm. Plugging these values into the expression for K, we get: $$1.3 \times 10^{5} = \frac{P_{H_2S}^2}{(0.103)^2 \times 0.417}$$
03

Solve for the equilibrium pressure of hydrogen sulfide (\(P_{H_2S}\))

Now, we just need to solve for \(P_{H_2S}\): $$P_{H_2S}^2 = 1.3 \times 10^{5} \times (0.103)^2 \times 0.417$$ $$P_{H_2S} = \sqrt{1.3 \times 10^{5} \times (0.103)^2 \times 0.417}$$ $$P_{H_2S} \approx 0.449 \text{ atm}$$ So, the equilibrium pressure of hydrogen sulfide is approximately 0.449 atm.

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Most popular questions from this chapter

Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{I}_{2}(g)+5 \mathrm{~F}_{2}(g) \rightleftharpoons 2 \mathrm{IF}_{5}(g)\) (b) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l)\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g)\) (d) \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

. At \(800 \mathrm{~K}\), hydrogen iodide can decompose into hydrogen and iodine gases. $$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{H}_{2}(g)$$ At this temperature, \(K=0.0169 .\) What are the partial pressures at equilibrium of the hydrogen and iodine if initially a sealed flask at \(800 \mathrm{~K}\) contains only HI at a pressure of \(0.200\) 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?

WEB Consider the following reaction at \(122^{\circ} \mathrm{C}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ (a) Write an equilibrium constant expression for the reaction and call the constant \(K_{1}\). (b) Write an equilibrium constant expression for the decomposition of one mole of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) and call the constant \(K_{2}\). (c) Relate \(K_{1}\) and \(K_{2}\).

Carbonylbromide (COBr_2) can be formed by combining carbon monoxide and bromine gas. $$\mathrm{CO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons \operatorname{COBr}_{2}(g)$$ When equilibrium is established at \(346 \mathrm{~K}\), the partial pressures (in atm) of \(\mathrm{COBr}_{2}\), \(\mathrm{CO}\), and \(\mathrm{Br}_{2}\) are \(0.12,1.00\), and \(0.65\), respectively. (a) What is \(K\) at \(346 \mathrm{~K} ?\) (b) Enough bromine condenses to decrease its partial pressure to \(0.50\) atm. What are the equilibrium partial pressures of all gases after equilibrium is re-established?
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