Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 40

Write the definition of equilibrium in terms of the general rate laws Rate \(_{\text {forward rxn }}=k_{\text {forward rxn }}[\text { Reactants }]^{\text {order }}\) and Rate \(_{\text {reverse rxn }}=k_{\text {reverse rxn }}[\text { Products }]^{\text {order }}\)

Short Answer

Expert verified
Equilibrium in terms of general rate laws can be defined when the forward and reverse reaction rates become equal, represented by the equation: \(k_{\text{forward rxn}}[\text{Reactants}]^{\text{order}} = k_{\text{reverse rxn}}[\text{Products}]^{\text{order}}\) Here, \(k_{\text{forward rxn}}\) and \(k_{\text{reverse rxn}}\) are the rate constants for forward and reverse reactions, and \(\text{order}\) represents the order of the reaction with respect to reactants and products.

Step by step solution

01

Forward Reaction Rate Law

The forward reaction rate law can be represented as: Rate\(_{\text{forward rxn}} = k_{\text{forward rxn}}[\text{Reactants}]^{\text{order}}\) Here, Rate\(_{\text{forward rxn}}\) represents the rate of the forward reaction, \(k_{\text{forward rxn}}\) is the rate constant for the forward reaction, \([\text{Reactants}]\) represents the concentration of the reactants, and \(\text{order}\) is the order of the reaction with respect to the reactants.
02

Reverse Reaction Rate Law

Similarly, the reverse reaction rate law can be represented as: Rate\(_{\text{reverse rxn}} = k_{\text{reverse rxn}}[\text{Products}]^{\text{order}}\) Here, Rate\(_{\text{reverse rxn}}\) represents the rate of the reverse reaction, \(k_{\text{reverse rx_{}{n}}}\) is the rate constant for the reverse reaction, \([\text{Products}]\) represents the concentration of the products, and \(\text{order}\) is the order of the reaction with respect to the products.
03

Equilibrium Condition

At equilibrium, the rates of the forward and reverse reactions become equal. This can be mathematically represented as: Rate\(_{\text{forward rxn}}\) = Rate\(_{\text{reverse rxn}}\)
04

Equilibrium Definition

Now, using the rate laws for forward and reverse reactions, we can define equilibrium as: \(k_{\text{forward rxn}}[\text{Reactants}]^{\text{order}} = k_{\text{reverse rxn}}[\text{Products}]^{\text{order}}\) This equation defines the condition for equilibrium in terms of the general rate laws for forward and reverse reactions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The saturation solubility of \(\mathrm{Ag}_{2} \mathrm{~S}\) at \(25^{\circ} \mathrm{C}\) is \(1.14 \times 10^{-17} \mathrm{M}\). What are the equilibrium concentrations of the cation and anion?

The equilibrium constant for the synthesis of methanol, $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}(g) $$ Methanol is \(4.3\) at \(250{ }^{\circ} \mathrm{C}\) and \(1.8\) at \(275^{\circ} \mathrm{C}\). (a) Does this reaction shift to the left or to the right when the reaction mixture is heated? Explain how you know. (b) Is this reaction endothermic or exothermic? Explain how you know. (c) Rewrite the equation for the reaction, including heat on the appropriate side.

At a given temperature, why is the ratio \(k_{\mathrm{f}} / k_{\mathrm{r}}\) constant for a given reaction?

Suppose you have a reaction with many reactants. When you write the equilibrium expression for the reaction, do the reactant concentrations all go in the numerator or in the denominator? What mathematical operation(s) should be used for these concentrations?

The solubility of silver acetate in water at \(20{ }^{\circ} \mathrm{C}\) is \(10.5 \mathrm{~g} / \mathrm{L}\) of solution. Calculate \(\mathrm{K}_{\mathrm{sp}}\) for silver acetate.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free