5. Consider the following reaction at equilibrium. What effect will each of the following changeshave on the amount of HCl present at equilibrium? (Increase, decrease, or remain the same)2Cl; (g) + 5 H2O(g) = 4HCI (g) + 6 O, (g) Endothermic reactiona. Decrease the temperatureb. Decrease the volume of the containerc. Remove H2O (g)d. Add a Catalyst

Answered: 5. Consider the following reaction at…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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5. Consider the following reaction at equilibrium. What effect will each of the following changes
have on the amount of HCl present at equilibrium? (Increase, decrease, or remain the same)
2Cl; (g) + 5 H2O(g) = 4HCI (g) + 6 O, (g) Endothermic reaction
a. Decrease the temperature
b. Decrease the volume of the container
c. Remove H2O (g)
d. Add a Catalyst

Expert Solution

Step 1

The effect of given changes on the amount of HCl at equilibrium in the given reaction is to be determined.

Endothermic reaction proceeds with gain of heat energy.

Catalyst increase the rate of forward and reverse reaction equally.

According to Le Chatelier’s Principle,

Increasing the partial pressure or concentration of a reactant shifts the equilibrium to product side (i.e. forward reaction).
Increasing the partial pressure or concentration of product shifts the equilibrium to reactant side (i.e. reverse reaction).
Decreasing the partial pressure or concentration of a reactant shifts the equilibrium toward the reactant side (i.e. reverse reaction).
Decreasing the partial pressure or concentration of product shifts the equilibrium to product side (i.e. forward reaction.
Increasing the total pressure of reaction shifts the equilibrium toward the side of the reaction equation with fewer moles of gases.
Decreasing the total pressure of reaction shifts the equilibrium toward the side of the reaction equation with more moles of gases.

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