Question

Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation

$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{\rightleftharpoons }\mathbf{C}\mathbf{+}\mathbf{D}$

a. You add more A to the flask. How does the concentration of each chemical compare with its original concentration after equilibrium is re-established? Justify your answer.

b. You have the original setup at equilibrium, and add more D to the flask. How does the concentration of each chemical compare with its original concentration after equilibrium is re-established? Justify your answer.

Short answer

(a) As reactant A is added more to the flask, the equilibrium will shift to the forward direction. This will lead …

Increase in concentration of A;

Decrease in concentration of B;

Increase in concentration of C;

Increase in concentration of D.

(b) As product D is added more to the flask, the equilibrium will shift to the backward direction. This will lead …

Increase in concentration of A;

Increase in concentration of B;

Decrease in concentration of C;

Increase in concentration of D.

Step-by-step solution

(a) Equilibrium constant for the reaction

For this reaction,

$\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$

You can write the equilibrium constant as

$\mathrm{K}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

Where, K = equilibrium constant;

$\left[\mathrm{A}\right]$= concentration of A at equilibrium;

$\left[\mathrm{B}\right]$= concentration of B at equilibrium;

$\left[\mathrm{C}\right]$= concentration of C at equilibrium;

$\left[\mathrm{D}\right]$= concentration of D at equilibrium;

La Chatelier’s principle

According to this principle, if a change is done on a system at equilibrium, the system will shift its equilibrium in such a direction to nullify the change.

Conclusion of (a)

As reactant A is added more to the flask, the equilibrium will shift to the forward direction. This will lead …

Increase in concentration of A;

Decrease in concentration of B;

Increase in concentration of C;

Increase in concentration of D.

Answer of subpart (b)

You need to predict about the concentrations of A, B, C and D; when D is added. Also you need to justify your answer.

You can answer this question with help of La Chatelier’s principle.

Equilibrium constant for the reaction

For this reaction,

$\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$

You can write the equilibrium constant as

$\mathrm{K}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

Where, K = equilibrium constant;

$\left[\mathrm{A}\right]$= concentration of A at equilibrium;

$\left[\mathrm{B}\right]$= concentration of B at equilibrium;

$\left[\mathrm{C}\right]$= concentration of C at equilibrium;

$\left[\mathrm{D}\right]$= concentration of D at equilibrium

La Chatelier’s principle

According to this principle, if a change is done on a system at equilibrium, the system will shift its equilibrium in such a direction to nullify the change.

Conclusion of (b)

As product D is added more to the flask, the equilibrium will shift to the backward direction. This will lead …

Increase in concentration of A;

Increase in concentration of B;

Decrease in concentration of C;

Increase in concentration of D.