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Chapter 12: Problem 24

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 \(1123 \mathrm{~K}\).

Short Answer

Expert verified
Answer: The equilibrium constant (K) for the given reaction at 1123 K is approximately 270.60.

Step by step solution

01

Write the balanced chemical equation

We are given that methane (CH4) and hydrogen sulfide (H2S) react to form carbon disulfide (CS2) and hydrogen gas (H2). Let's write this reaction and balance it: CH4 + H2S → CS2 + H2 To balance the reaction, we can see that we need to add a 2 in front of H2: CH4 + H2S → CS2 + 2H2 Now, the reaction is balanced.
02

Write the expression for the equilibrium constant (K)

The equilibrium constant, K, is the ratio of the concentrations of the products and the reactants raised to their stoichiometric coefficients in the balanced chemical equation. For the reaction: CH4 + H2S → CS2 + 2H2 We can write the K expression as: K = [CS2] * [H2]^2 / ([CH4] * [H2S])
03

Calculate K using the given equilibrium concentrations

We have the equilibrium concentrations of each substance: [CH4] = 0.00142 M, [H2S] = 6.14 x 10^{-4} M, [CS2] = 0.00266 M, and [H2] = 0.00943 M. Plug these values into the K expression: K = (0.00266) * (0.00943)^2 / (0.00142 * (6.14 * 10^{-4})) Now, calculate K: K ≈ (0.00266 * 0.00008892) / (0.00142 * 0.000614) K ≈ 0.23602752 / 0.00087188 K ≈ 270.60
04

Report the equilibrium constant K

The equilibrium constant K for the given reaction at 1123 K is approximately 270.60.

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Most popular questions from this chapter

Hydrogen cyanide, a highly toxic gas, can decompose to cyanogen and hydrogen gases, $$2 \mathrm{HCN}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{~N}_{2}(g)+\mathrm{H}_{2}(g)$$ At a certain temperature, \(K\) for this decomposition is \(0.17\). What are the partial pressures of all gases at equilibrium if initially the partial pressures are \(P_{\mathrm{C}_{2} \mathrm{~N}_{2}}=P_{\mathrm{H}_{2}}=0.32 \mathrm{~atm}, P_{\mathrm{HCN}}=0.45 \mathrm{~atm} ?\)

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?

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

Carbonyl fluoride, \(\mathrm{COF}_{2}\), is an important intermediate for organic fluorine compounds. It can be prepared by the following reaction: $$\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CF}_{4}(g) \rightleftharpoons 2 \mathrm{COF}_{2}(g)$$ At \(1000^{\circ} \mathrm{C}, K\) for this reaction is \(0.50 .\) What are the partial pressures of all the gases at equilibrium when the initial partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CF}_{4}\) are \(0.713 \mathrm{~atm} ?\)

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{1}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{C}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{3} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)
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