Chapter 6, Problem 120IAE

Interpretation Introduction

(a)

Interpretation:

To determine the percentage of molecules whose speed is 0 ms-1.

Concept introduction:

Due to the larger number of molecules, the speed of the each of them cannot be calculated or known at the same time. The equation shown below can be used as it describes the distribution of speeds in molecules in a given sample of gas.

$\text{F(u)}=4\pi {\left(\frac{M}{2\pi \text{RT}}\right)}^{\frac{3}{2}}\times {\text{u}}^{\text{2}}{\text{e}}^{\frac{{\text{-Mu}}^{\text{2}}}{\text{2RT}}}$

Interpretation Introduction

(b)

Interpretation:

To determine the percentage of molecules whose speed is 500 ms-1.

Concept introduction:

The equilibrium constant K is used to express the relation between the reactants and the products of a given reaction at equilibrium with respect to some specific unit. The relation between Gibbs Free energy and the equilibrium constant is given by the equation:

${\text{ΔG}}^{\text{ο}}{\text{= -RTlnK}}_{\text{eq}}$

Interpretation Introduction

(c)

Interpretation:

To determine the percentage of molecules whose speed is 1000 ms-1.

Concept introduction:

The equilibrium constant K is used to express the relation between the reactants and the products of a given reaction at equilibrium with respect to some specific unit. The relation between Gibbs Free energy and the equilibrium constant is given by the equation:

${\text{ΔG}}^{\text{ο}}{\text{= -RTlnK}}_{\text{eq}}$

Interpretation Introduction

(d)

Interpretation:

To determine the percentage of molecules whose speed is 1500 ms-1.

Concept introduction:

The equilibrium constant K is used to express the relation between the reactants and the products of a given reaction at equilibrium with respect to some specific unit. The relation between Gibbs Free energy and the equilibrium constant is given by the equation:

${\text{ΔG}}^{\text{ο}}{\text{= -RTlnK}}_{\text{eq}}$

Interpretation Introduction

(e)

Interpretation:

To determine the percentage of molecules whose speed is 2000 ms-1.

Concept introduction:

The equilibrium constant K is used to express the relation between the reactants and the products of a given reaction at equilibrium with respect to some specific unit. The relation between Gibbs Free energy and the equilibrium constant is given by the equation:

${\text{ΔG}}^{\text{ο}}{\text{= -RTlnK}}_{\text{eq}}$

Interpretation Introduction

(f)

Interpretation:

To determine the percentage of molecules whose speed is 2500 ms-1.

Concept introduction:

The equilibrium constant K is used to express the relation between the reactants and the products of a given reaction at equilibrium with respect to some specific unit. The relation between Gibbs Free energy and the equilibrium constant is given by the equation:

${\text{ΔG}}^{\text{ο}}{\text{= -RTlnK}}_{\text{eq}}$

Interpretation Introduction

(g)

Interpretation:

To determine the percentage of molecules whose speed is 3500 ms-1 and graph should be drawn for obtained results.

Concept introduction:

The equilibrium constant K is used to express the relation between the reactants and the products of a given reaction at equilibrium with respect to some specific unit. The relation between Gibbs Free energy and the equilibrium constant is given by the equation:

${\text{ΔG}}^{\text{ο}}{\text{= -RTlnK}}_{\text{eq}}$