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Chapter 14: Problem 40

The equilibrium-constant expression for a reaction is $$ K_{c}=\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}{[\mathrm{NO}]^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}} $$ What is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

Short Answer

Expert verified
The new expression is the inverse and square of the initial: \[ \left( \frac{[\text{NH}_3]^2 [\text{O}_2]^{2.5}}{\left[\text{NO}\right]^2\left[\text{H}_2\text{O}\right]^3} \right)^2 \].

Step by step solution

01

Understand the Initial Expression

The given equilibrium-constant expression is for a reaction, indicating how the concentration of products and reactants relate when the reaction is at equilibrium. It is written as:\[ K_{c} = \frac{[\text{NH}_3]^4 [\text{O}_2]^5}{[\text{NO}]^4 [\text{H}_2\text{O}]^6} \] This relates to a balanced chemical equation which is not provided here.
02

Transform the Chemical Equation

When an equation is halved, each stoichiometric coefficient is divided by 2. This makes the reaction more simplified. Afterward, reverse the reaction, which means the products and reactants swap places.
03

Determine the New Equilibrium Expression

If the stoichiometry of a reaction is changed by a factor, the equilibrium constant expression will be affected accordingly. For halving, each term in the expression must be squared because the coefficients are halved from the original reaction:\[ \frac{1}{2} \text{Coefficient} \rightarrow K_c^2 \] Since the reaction is reversed, the expression becomes the reciprocal of the squared version:\[ K_{c,\text{new}} = \left( \frac{\left[\text{NO}\right]^{4/2}\left[\text{H}_2\text{O}\right]^{6/2}}{\left[\text{NH}_3\right]^{4/2}\left[\text{O}_2\right]^{5/2}} \right)^2 \]Simplifying, we find:\[ K_{c,\text{new}} = \left( \frac{\left[\text{NO}\right]^2\left[\text{H}_2\text{O}\right]^3}{\left[\text{NH}_3\right]^2\left[\text{O}_2\right]^{2.5}} \right)^2 \] Finally, because the equation is reversed, we take the reciprocal:\[ K_{c,\text{final}} = \frac{1}{K_{c,\text{new}}} = \left( \frac{\left[\text{NH}_3\right]^2\left[\text{O}_2\right]^{2.5}}{\left[\text{NO}\right]^2\left[\text{H}_2\text{O}\right]^3} \right)^2 \]

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state of a chemical reaction where the concentrations of reactants and products remain constant over time.
This happens because the rate of the forward reaction (reactants turning into products) is equal to the rate of the reverse reaction (products turning back into reactants).
At this point, no net change occurs in the concentrations of reactants and products. In a balanced chemical equation, we can use the equilibrium constant ( \(K_{c}\) ) to express the ratio of the concentrations of products to reactants.
This ratio remains constant at equilibrium and is specific to a given reaction at a particular temperature.
  • An equilibrium constant greater than one ( \(K_{c} > 1\) ) typically suggests that, at equilibrium, the products are favored.
  • Conversely, an equilibrium constant less than one ( \(K_{c} < 1\) ) indicates a reaction that favors the reactants.
Understanding chemical equilibrium helps chemists predict how changes in concentration, temperature, or pressure will affect a reaction.
Stoichiometry
Stoichiometry is the part of chemistry that involves the calculation of reactants and products in chemical reactions.
It uses the coefficients from the balanced chemical equation to relate quantities of substances. When deriving an equilibrium-constant expression from a chemical equation, stoichiometry plays a crucial role.
The stoichiometric coefficients become the exponents in the equilibrium-constant expression.
  • In the equation \([A]^{a}[B]^{b} \rightarrow [C]^{c}[D]^{d} \), the equilibrium constant expression is given by \(\ K_{c} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).
  • Halving a reaction means reducing all stoichiometric coefficients by half, impacting the equilibrium constant by the square of the changes.
This understanding allows you to predict how altering the stoichiometry of a reaction, such as halving or doubling it, affects the equilibrium constants.
Reaction Kinetics
Reaction kinetics focuses on the speed or rate at which chemical reactions occur. It offers insights into how various factors, such as temperature and concentration, influence reaction rates. When analyzing equilibrium reactions, it's important to understand that kinetics and equilibrium are closely related.
In a state of equilibrium, the reaction rates of the forward and reverse reactions are equal.
  • Factors influencing kinetics, such as catalysts or temperature changes, can shift equilibrium in accordance with Le Chatelier鈥檚 principle.
  • While stoichiometry tells us the ratio of reactants to products at equilibrium, kinetics explains how quickly a reaction reaches that point.
Understanding reaction kinetics helps predict how changes in conditions will influence both the rate at which equilibrium is established and the position of that equilibrium.

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Most popular questions from this chapter

Nitrogen monoxide, NO, reacts with bromine, \(\mathrm{Br}_{2}\), to give nitrosyl bromide, NOBr. $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ A sample of \(0.0524\) mol \(\mathrm{NO}\) with \(0.0262 \mathrm{~mol} \mathrm{Br}_{2}\) gives an equilibrium mixture containing \(0.0311\) mol NOBr. What is the composition of the equilibrium mixture?

A 2.0-L reaction flask initially contains \(0.10 \mathrm{~mol} \mathrm{CO}\), \(0.20 \mathrm{~mol} \mathrm{H}_{2}\), and \(0.50 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\) (methanol). If this mixture is brought in contact with a zinc oxide-chromium(III) oxide catalyst, the equilibrium $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ is attained. The equilibrium constant \(K_{c}\) for this reaction at \(300^{\circ} \mathrm{C}\) is \(1.1 \times 10^{-2}\). What is the direction of reaction (forward or reverse) as the mixture attains equilibrium?

List the possible ways in which you can alter the equilibrium composition of a reaction mixture.

Would either of the following reactions go almost completely to product at equilibrium? a. \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ; K_{c}=6.5 \times 10^{113}\) b. \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{c}=3 \times 10^{-17}\)

Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) (from air). $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{c}\) is \(0.0025\) at \(2127^{\circ} \mathrm{C}\). If an equilibrium mixture at this temperature contains \(0.023 \mathrm{~mol} \mathrm{~N}_{2}\) and \(0.031 \mathrm{~mol} \mathrm{O}_{2}\) per liter, what is the concentration of NO?
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