Chapter 15, Problem 96IAE

Interpretation Introduction

(a)

Interpretation:

Reaction quotient of the reactions should be shown.

Concept introduction:

Using balanced chemical equation, reaction quotient for a reaction can be written in terms of activity and thee corresponding coefficient.

At the equilibrium both Q and K has the same value.

Interpretation Introduction

(b)

Interpretation:

The value of ${\text{Q}}_{\text{P}}$ for each mixture should be calculated.

Concept introduction:

Using balanced chemical equation, reaction quotient for a reaction can be written in terms of activity and thee corresponding coefficient.

At the equilibrium both Q and K has the same value.

For the given reaction, the expression for the ${\text{Q}}_{\text{P}}$ is written as follows:

${\text{Q}}_{\text{P}}\text{=}\frac{{\text{X}}^{\text{2}}{}_{{\text{NH}}_{\text{3}}}}{{\text{X}}_{{\text{N}}_{\text{2}}}{\text{X}}^{\text{3}}{}_{{\text{H}}_{\text{2}}}}\times \frac{1}{{\text{P}}^{\text{2}}}$

Interpretation Introduction

(c)

Interpretation:

Net reaction of the A and B mixture should be explained using Q_c and K_c.

Concept introduction:

For the given reaction, the equation for Qp is as follows:

${\text{Q}}_{\text{P}}\text{=}\frac{{\text{X}}^{\text{2}}{}_{{\text{NH}}_{\text{3}}}}{{\text{X}}_{{\text{N}}_{\text{2}}}{\text{X}}^{\text{3}}{}_{{\text{H}}_{\text{2}}}}\times \frac{1}{{\text{P}}^{\text{2}}}$

At the equilibrium Q_p is equal to the K_p.

If Q_p is larger than the K_p, then the reaction is favored to left side.

If Q_p is lower than the K_p, then the reaction is favored to right side.