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Chapter 14: Problem 79

At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)$$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right]\)

Short Answer

Expert verified
The equilibrium concentration of sulfur, \([\mathrm{S}_{2}]\), is calculated to be approximately \(1.83 \times 10^{-2} \mathrm{M}\).

Step by step solution

01

Understand the Equation

The equation describes the decomposition of hydrogen sulphide into hydrogen and sulfur in a gaseous state. The stoichiometric coefficients in the balanced chemical equation are crucial to understand how the molar ratios of the reactants and products relate. In the balanced equation, 2 moles of hydrogen sulphide yield 2 moles of hydrogen and 1 mole of sulfur.
02

Apply the Equilibrium Constant Formula

The formula \(K_{c} = \frac{[\mathrm{H}_{2}]^{2}[\mathrm{S}_{2}]}{[\mathrm{H}_{2}\mathrm{S}]^{2}}\) is used to express the equilibrium constant. This formula is derived from the balanced chemical equation. The concentration of each product is raised to the power of its stoichiometric coefficient in the balanced equation and then multiplied together for the numerator. The same is done for the reactants for the denominator.
03

Extract Information and Substitute into the Formula

With the formula in mind, the given values can be substituted to calculate for \([\mathrm{S}_{2}]\):\[\begin{align*}2.25 \times 10^{-4} = \frac{(1.50 \times 10^{-3} \mathrm{M})^{2}\times [\mathrm{S}_{2}]}{(4.84 \times 10^{-3} \mathrm{M})^{2}}\end{align*}\]This equation can be simplified and then solved for \([\mathrm{S}_{2}]\).
04

Solve for \([\mathrm{S}_{2}]\)

Upon simplification of the equation from Step 3, \([\mathrm{S}_{2}]\) becomes the subject of the formula:\[\begin{align*}[\mathrm{S}_{2}] = 2.25 \times 10^{-4} \times \frac{(4.84 \times 10^{-3})^{2}}{(1.50 \times 10^{-3})^{2}}\end{align*}\]This equation can be solved to find the equilibrium concentration of sulfur.

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Most popular questions from this chapter

Baking soda (sodium bicarbonate) undergoes thermal decomposition as follows: $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

A student placed a few ice cubes in a drinking glass with water. A few minutes later she noticed that some of the ice cubes were fused together. Explain what happened.

Consider the following reaction, which takes place in a single elementary step: $$2 \mathrm{~A}+\mathrm{B} \underset{k_{-1}}{\frac{k_{1}}{\longrightarrow}} \mathrm{A}_{2} \mathrm{~B}$$ If the equilibrium constant \(K_{\mathrm{c}}\) is 12.6 at a certain temperature and if \(k_{\mathrm{r}}=5.1 \times 10^{-2} \mathrm{~s}^{-1},\) calculate the value of \(k_{\mathrm{f}}\).

The equilibrium constant \(K_{P}\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C}\). The initial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in a mixture are 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures \((1.112 \mathrm{~atm}) ?\)

Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)$$ What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system; (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system; (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.
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