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Chapter 14: Problem 65

Determine the initial and equilibrium concentrations of HI if the initial concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) are both \(0.16 M\) and their equilibrium concentrations are both \(0.072 M\) at \(430^{\circ} \mathrm{C}\). The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) is 54.2 at \(430^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The initial concentration of HI is 0 M and the equilibrium concentration of HI is 0.176 M.

Step by step solution

01

Analyze the Reaction and Write the Equilibrium Expression

The reaction given is \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\). The equilibrium constant expression for this reaction is \(K_{c} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}\). This is derived from the balanced chemical equation. Because the coefficient of HI is 2, its concentration is squared in the equilibrium equation.
02

Calculate the Change in Concentration

Both the initial concentrations of H2 and I2 decrease by 0.088 M, therefore, the concentration of HI would increase by 2*0.088 M = 0.176 M because 2 moles of HI are produced from 1 mole each of H2 and I2. Hence, for every 0.088 M used up, 0.176 M is produced.
03

Determine the Equilibrium Concentrations

Adding the change to the starting concentration gives the equilibrium concentration of HI as 0 M + 0.176 M = 0.176 M.
04

Check With the Equilibrium Constant

Substitute all the equilibrium concentrations into the equilibrium expression obtained in step 1 and solve for the equilibrium constant. If the calculated value matches the given value of Kc, then the equilibrium concentration is correct. Here, \(K_{c} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}= \dfrac{(0.176)^2 }{(0.072) \times (0.072)} = 54.2\). As these values match exactly, it can be confirmed that the equilibrium concentrations are correct!

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Most popular questions from this chapter

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=260 \mathrm{~kJ} /\mathrm{mol}\end{array}$$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\), in the presence of air, to convert the remaining methane to hydrogen: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

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.

List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it?

Consider the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If nitrosyl bromide, \(\mathrm{NOBr},\) is 34 percent dissociated at \(25^{\circ} \mathrm{C}\) and the total pressure is 0.25 atm, calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the dissociation at this temperature.

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 \(k_{\mathrm{f}}\).
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