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 \(K_{\mathrm{c}}\) for the decomposition of phosgene, \(\mathrm{COCl}_{2}\), is \(4.63 \times 10^{-3}\) at \(527^{\circ} \mathrm{C}\) : $$\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$ Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm.

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 1.2 at \(375^{\circ} \mathrm{C}\). $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) \(\rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

What is the law of mass action?

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 \(430^{\circ} \mathrm{C}\)

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol} .\) If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1},\) what is \(k_{1}\) at \(298 \mathrm{~K} ?\) The equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).

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