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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 reaction \( \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) \) at the same temperature is \( \frac{1}{4.17 \times 10^{-34}} \).

Step by step solution

01

Understand the relationship between the equilibrium constants for forward and reverse reactions

The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. This is because for a reaction in equilibrium, both the forward and reverse reactions are taking place, and if the reaction quotient for the forward reaction is \(K_c\), then the reaction quotient for the reverse reaction is always \( \frac{1}{K_c} \).
02

Apply the reciprocal rule to find the equilibrium constant for the reverse reaction

Given that the equilibrium constant \(K_c\) for the forward reaction is \(4.17 \times 10^{-34}\), the equilibrium constant for the reverse reaction will be calculated as \( \frac{1}{4.17 \times 10^{-34}} \).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \( K_c \) or \( K_{eq} \), serves as a key indicator of the extent to which a chemical reaction proceeds before reaching the state of equilibrium. It is defined as the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients in a balanced chemical equation, when the reaction has reached equilibrium at a constant temperature.

For a generic reaction where \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant would be expressed as:\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
It’s essential to understand that \( K_c \) is constant only at a given temperature; any change in temperature could affect the value of the equilibrium constant, thus also affecting the position of the equilibrium.
Reciprocal Rule in Chemistry
The reciprocal rule in chemistry is an important concept when dealing with reversible reactions and their equilibrium constants. This rule states that if a reaction in one direction has an equilibrium constant \( K_c \), the equilibrium constant for the reaction in the reverse direction is \( 1/K_c \).

This is attributed to the fact that the reaction's dynamics are symmetrical; the forward process's propensity to create products at equilibrium is equally balanced by the reverse process's tendency to revert to reactants. Mathematically, if the forward reaction is represented as \( A \rightleftharpoons B \) with equilibrium constant \( K_{c1} \), the reverse \( B \rightleftharpoons A \) would have an equilibrium constant of \( K_{c2} = 1/K_{c1} \).
Chemical Equilibrium
Chemical equilibrium occurs in a reversible chemical reaction when the rates of the forward and reverse reactions equalize, resulting in no overall change in the concentrations of reactants and products over time. It is important to note that even though the macroscopic properties are not changing, the reaction remains dynamic; reactants and products continually form, albeit at the same rate.

At equilibrium, the reaction has reached a state of balance, but this does not mean the reactants and products are present in equal concentrations—rather, their ratios are stable. The concept of equilibrium plays a fundamental role in predicting product formation and reactant consumption in chemical reactions.
Reaction Quotient
The reaction quotient, denoted as \( Q \), is a numerical value that shows the relative amounts of products and reactants present during a reaction at any given point in time before equilibrium is reached. It is calculated using the same formula as for the equilibrium constant, but with the current concentrations rather than the equilibrium concentrations.

The value of \( Q \) in comparison to the equilibrium constant \( K_c \) allows prediction of the system’s direction to reach equilibrium. If \( Q < K_c \), the system tends to form more products, moving forward. If \( Q > K_c \), the system tends to form more reactants, shifting the reaction backward. When \( Q = K_c \), the system is already at equilibrium.

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Most popular questions from this chapter

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the change in concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) with time for these situations: (a) Initially only A is present; (b) initially only B is present; (c) initially both \(\mathrm{A}\) and \(\mathrm{B}\) are present (with \(\mathrm{A}\) in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Explain the difference between physical equilibrium and chemical equilibrium. Give two examples of each.

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are: \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.491 \mathrm{M}\). The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(4.63 \times\) \(10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color.]

A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M},\left[\mathrm{N}_{2}\right]=0.11 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=1.91 \mathrm{M}\). Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)

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