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Chapter 14: Problem 15

The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)$$ is \(4.17 \times 10^{-34}\) at \(25^{\circ} \mathrm{C} .\) What is the equilibrium constant for the reaction $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$$ at the same temperature?

Short Answer

Expert verified
The equilibrium constant for the reverse reaction at the same temperature is \( 2.40 \times 10^{33} \).

Step by step solution

01

Understand the relationship between reverse and forward reactions

From the principle of Chemical Equilibrium, if a reaction, say A yields B with an equilibrium constant K1, its reverse reaction i.e. B yields A will have the equilibrium constant as \( K_{2} = 1/K_{1} \)
02

Apply this principle to given problem

Here, the equilibrium constant \( K_{c} \) of the forward reaction is given as \(4.17 \times 10^{-34}\). So applying the principle for the equilibrium constant of reverse reaction, which is$$ K_{c(reverse)} = 1 / K_{c(forward)} $$ we substitute the given \( K_{c} \) into the equation.
03

Solve for the equilibrium constant of the reverse reaction

Substituting the given \( K_{c} \) into the equation, we have:$$ K_{c(reverse)} = 1 / (4.17 \times 10^{-34}) $$ Solving this gives us the \( K_{c} \) of the reverse reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. As a system reaches equilibrium, both the forward and reverse reactions continue to occur, but the concentrations of the reactants and products remain constant. By achieving this state, a dynamic balance is established where the amount of reactants turning into products is equal to the amount of products reverting back into reactants.

Understanding chemical equilibrium is akin to watching a video on pause: although no noticeable change occurs over time, there's plenty of action if you 'play' the scene. This analogously represents the molecular action at equilibrium; substances interconvert but the overall picture remains the same.
Reverse Reaction
The term 'reverse reaction' is used to describe the process where the products of a reaction reform the original reactants. It quite literally is the reverse of the direction considered in the 'forward reaction'. The concept of reversible reactions is fundamental in understanding chemical equilibrium because it's through these opposing processes that equilibrium is established.

If we think of a chemical reaction as a two-way street, the reverse reaction is traffic flowing back to the starting point. In the provided exercise, the reverse reaction is represented as the formation of HCl gas from H2 and Cl2 gases. When these reactions occur at the same rate as their forward counterparts, no net change is observed - the traffic on both sides is equal, and our chemical 'road' reaches a steady state.
Forward Reaction
In contrast to the reverse reaction, the 'forward reaction' is the process in which reactants convert to products. It's the direction that is typically followed when writing out a chemical equation. During this process, reactants are consumed to form products until a state of chemical equilibrium is reached (if the reaction is reversible).

Continuing with the previous analogy, if the reverse reaction is traffic back to the starting point, then the forward reaction is traffic heading towards the destination. For the exercise in question, the forward reaction is described as the breakdown of hydrogen chloride (HCl) gas into hydrogen (H2) and chlorine (Cl2) gases.
Equilibrium Principles
Equilibrium principles, or the laws that govern the state of chemical equilibrium, are the rules that help us understand and predict the behavior of reversible reactions. The most fundamental principle is that at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This principle applies to all reversible reactions, regardless of their complexity.

Another key principle relates to the equilibrium constant, denoted by Kc. This value is a measure of the ratio of the concentration of products to the concentration of reactants at equilibrium. It is a constant for a given reaction at a specific temperature. For the exercise, by understanding that the equilibrium constant of the reverse reaction (the formation of HCl gas) is the reciprocal of the equilibrium constant of the forward reaction (the breakdown of HCl gas), students can correctly solve for the equilibrium state of a system.

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Most popular questions from this chapter

The equilibrium constant \(\left(K_{P}\right)\) for the formation of the air pollutant nitric oxide (NO) in an automobile engine at \(530^{\circ} \mathrm{C}\) is \(2.9 \times 10^{-11}\) : $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.0 atm and 0.012 atm, respectively. (b) Repeat the calculation for atmospheric conditions where the partial pressures of nitrogen and oxygen are 0.78 atm and 0.21 atm and the temperature is \(25^{\circ} \mathrm{C}\). (The \(K_{P}\) for the reaction is \(4.0 \times 10^{-31}\) at this temperature.) (c) Is the formation of NO endothermic or exothermic? (d) What natural phenomenon promotes the formation of NO? Why?

Consider the reaction between \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in a closed container: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ Initially, 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(\alpha\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(\alpha\) and \(P\), the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium due to an increase in \(P ?\) Does your prediction agree with Le Ch芒telier's principle?

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?

Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]

The equilibrium constant \(K_{P}\) for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures \(0.177 \mathrm{~atm},\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.
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