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If two reactions sum to an overall reaction, and the equilibrium constants for the two reactions are \(K_{1}\) and \(K_{2},\) what is the equilibrium constant for the overall reaction?

Short Answer

Expert verified
The equilibrium constant for the overall reaction is the product of the two equilibrium constants, \(K_{overall} = K_{1} \times K_{2}\).

Step by step solution

01

Understand the relationship between equilibrium constants of reactions in series

When two chemical reactions sum to form an overall reaction, the equilibrium constants for these reactions are related multiplicatively. If the two reactions are: Reaction 1 with equilibrium constant \(K_1\) and Reaction 2 with equilibrium constant \(K_2\), the equilibrium constant for the overall reaction \(K_{overall}\) is the product of \(K_1\) and \(K_2\).
02

Derive the equilibrium constant for the overall reaction

Since the overall reaction is the sum of Reaction 1 and Reaction 2, its equilibrium constant is mathematically represented by the product \(K_1 \times K_2\). This is because the reaction quotients of the two reactions multiply together when the reactions are added, according to the properties of equilibrium systems.
03

State the expression for the overall equilibrium constant

The equilibrium constant for the overall reaction is thus given by the expression \(K_{overall} = K_1 \times K_2\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that describes a state in which the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of the reactants and products over time. In a reaction such as \( A + B \leftrightarrow C + D \), the system reaches equilibrium when the rate of the reaction converting \( A \) and \( B \) into \( C \) and \( D \) is equal to the rate of the reverse reaction turning \( C \) and \( D \) back into \( A \) and \( B \). At this point, the ratio of the products to the reactants remains constant.

It's crucial to understand that equilibrium does not imply that the reactants and products are present in equal amounts. Instead, it indicates that their ratios are stable. This balanced state is represented by an equilibrium constant \(K\), which is a numerical value reflecting the ratio of the concentration of the products to the reactants, each raised to the power of their coefficients from the balanced chemical equation.
Reaction Quotient
The reaction quotient, designated as \(Q\), is a ratio that compares the current concentrations of products and reactants at any point before the system has reached equilibrium. It serves as a predictor to determine the direction in which a reaction will proceed to achieve equilibrium. Essentially, it uses the same formula as the equilibrium constant but with the initial concentrations, rather than the equilibrium concentrations.

When comparing \(Q\) to the equilibrium constant \(K\), if \(Q < K\), the forward reaction will be favored until equilibrium is reached. Conversely, if \(Q > K\), the reverse reaction will be favored. When \(Q = K\), the system is already at equilibrium. Monitoring the reaction quotient helps chemists to understand the progress of a reaction and predict its behavior over time.
Equilibrium Systems
Equilibrium systems encapsulate all the components and conditions involved in an equilibrium state. These systems are dynamic, meaning that the forward and reverse reactions continue to occur, but they do so in such a way that the observable properties of the system, like concentrations, temperature, and pressure, remain constant if the system is closed and unchanging.

An important aspect of equilibrium systems is their response to changes in conditions, which is described by Le Chatelier’s Principle. This principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to counteract the change and re-establish equilibrium. Understanding equilibrium systems is essential for manipulating reaction conditions to favor the production of desired products in chemical manufacturing processes.
Equilibrium Constants Relationship
The relationship between equilibrium constants for multiple reactions can be understood by considering the combined reactions as a system. When two reactions with known constants \(K_1\) and \(K_2\) combine to form an overall reaction, their constants are not added; instead, they are multiplied to find the equilibrium constant for the overall reaction, \(K_{overall}\).

Thus, if you have two reactions occurring sequentially:\[\text{Reaction 1: } A \rightarrow B \quad (K_1)\]
\[\text{Reaction 2: } B \rightarrow C \quad (K_2)\]
The overall reaction from \(A \rightarrow C\) has its equilibrium constant determined by the product of the individual ones: \[K_{overall} = K_1 \times K_2\]. This multiplicative relationship is derived from the properties of chemical equilibrium and can be explained using the concept of reaction quotients, as each step towards equilibrium is an independent event that multiplies together.

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Most popular questions from this chapter

What is the effect of a temperature change on a chemical reaction ini- tially at equilibrium? How does the effect differ for an exothermic reac- tion compared to an endothermic one?

When this reaction comes to equilibrium, will the concentrations reactants or products be greater? Does the answer to this question pend on the initial concentrations of the reactants and products? \begin{equation}\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons 2 \mathrm{C}(g) \quad K_{\mathrm{c}}=1.4 \times 10^{-5}\end{equation}

Write an expression for the equilibrium constant of each chemical equation. \begin{equation} \begin{array}{l}{\text { a. } \operatorname{SbCl}_{5}(g) \Longrightarrow \operatorname{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g)} \\ {\text { b. } 2 \mathrm{BrNO}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)} \end{array}\end{equation} \begin{equation} \begin{array}{l}{\text { c. } \operatorname{CH}_{4}(g) \Longrightarrow \operatorname2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{CS}_{2}(g)} \\ {\text { d. } 2 \mathrm{CO}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm2 \mathrm{CO}_{2}(g)} \end{array}\end{equation}

What happens to the value of the equilibrium constant for a reaction if the reaction equation is reversed? Multiplied by a constant?

Consider the reaction: $$\begin{array}{c}{\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)} \\\ {K_{\mathrm{c}}=1.8 \times 10^{-5} \mathrm{at} 25^{\circ} \mathrm{C}}\end{array}$$ If a solution initially contains \(0.210 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},\) what is the equi- librium concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) at \(25^{\circ} \mathrm{C}\) ?

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