Chapter 17, Problem 17.80P

(a)

Interpretation Introduction

Interpretation:

The given reaction is

  ${\text{X}}_{\text{2(g)}}{\text{+ Y}}_{\text{2(g)}}\rightleftharpoons {\text{2XY}}_{\text{(g)}}$

The reaction quotient, $Q$, for the given reaction has to be written.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by $Q$.  For a reversible reaction of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ to NO}}_{\text{2}}$, the reaction quotient is represented as

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}\left(\text{g}\right)}\rightleftharpoons {\text{2NO}}_{\text{2(g)}}\\ \\ Q\text{ = }\frac{{\left({\text{NO}}_{\text{2}}\right)}^{\text{2}}}{\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)}\end{array}$

Explanation of Solution

The given reaction is

  ${\text{X}}_{\text{2(g)}}{\text{+ Y}}_{\text{2(g)}}\rightleftharpoons {\text{2XY}}_{\text{(g)}}$

The reaction quotient for the reaction can be represented as

  $Q\text{ = }\frac{{\text{(XY)}}^{\text{2}}}{{\text{(X}}_{\text{2}}{\text{)(Y}}_{\text{2}}\text{)}}$

(b)

Interpretation Introduction

Interpretation:

The given filmstrip contains five molecular scenes of a gaseous mixture as it reaches equilibrium over time.  In the filmstrip, $\text{X}$ is purple and $\text{Y}$ is orange.

Figure 1

If each particle represents $\text{0}\text{.1 mol}$, $Q$ has to be found for each scene.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by $Q$.  For a reversible reaction of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ to NO}}_{\text{2}}$, the reaction quotient is represented as

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}\left(\text{g}\right)}\rightleftharpoons {\text{2NO}}_{\text{2(g)}}\\ \\ Q\text{ = }\frac{{\left({\text{NO}}_{\text{2}}\right)}^{\text{2}}}{\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)}\end{array}$

Explanation of Solution

Scene A:

$\text{X}$ Molecules are $\text{4}$, $\text{Y}$ molecules are $\text{4}$ and $\text{XY}$ molecules are absent.  As each particle represents $\text{0}\text{.1 mol}$,

  $\begin{array}{l}\text{X = 4×0}\text{.1 = 0}\text{.4 mol}\\ \\ \text{Y = 4×0}\text{.1 = 0}\text{.4 mol}\end{array}$

The $Q$ can be represented as

  $Q\text{ = }\frac{{\text{(0)}}^{\text{2}}}{\text{(0}\text{.4)(0}\text{.4)}}\text{ = 0}$

Scene B:

$\text{X}$ Molecules are $\text{2}$, $\text{Y}$ molecules are $\text{2}$ and $\text{XY}$ molecules are $\text{4}$.  As each particle represents $\text{0}\text{.1 mol}$,

  $\begin{array}{l}\text{X }\text{= 2×0}\text{.1 = 0}\text{.2 mol}\\ \\ \text{Y }\text{= 2×0}\text{.1 = 0}\text{.2 mol}\\ \\ \text{XY}\text{=4×0}\text{.1 = 0}\text{.4mol}\end{array}$

The $Q$ can be represented as

  $Q\text{ = }\frac{{\text{(0}\text{.4)}}^{\text{2}}}{\text{(0}\text{.2)(0}\text{.2)}}\text{ = 4}$

Scene C, D, E:

$\text{X}$ Molecules are $1$, $\text{Y}$ molecules are $1$ and $\text{XY}$ molecules are $6$.  As each particle represents $\text{0}\text{.1 mol}$,

  $\begin{array}{l}\text{X}\text{= 1×0}\text{.1 = 0}\text{.1 mol}\\ \\ \text{Y}\text{= 1×0}\text{.1 = 0}\text{.1 mol}\\ \\ \text{XY}\text{=6×0}\text{.1 =0}\text{.6mol}\end{array}$

The $Q$ can be represented as

  $Q\text{ = }\frac{{\text{(0}\text{.6)}}^{\text{2}}}{\text{(0}\text{.1)(0}\text{.1)}}{\text{ = 36 = 4×10}}^{\text{1}}\text{ mol}$

(c)

Interpretation Introduction

Interpretation:

The given reaction is

  ${\text{X}}_{\text{2(g)}}{\text{+ Y}}_{\text{2(g)}}\rightleftharpoons {\text{2XY}}_{\text{(g)}}$

If $\text{K > 1}$, the time progressing to the right or to the left has to be explained.

Concept Introduction:

At equilibrium, $Q\text{ = K}$.

If $Q\text{ = K}$, the mixture is at equilibrium.

If $Q\text{ > K}$, the reaction proceeds towards reactants.

If $Q\text{ < K}$, the reaction proceeds towards products.

Explanation of Solution

The given condition is $\text{K > 1}$.

Scene-A:

The value of $Q$ is $\text{0}$.  It is less than $\text{1}$.  Hence, $Q\text{ < K}$.

Scene-B:

The value of $Q$ is $4$.  It is greater than $\text{1}$.  Hence, $Q\text{ = K}$.

Scene-C, D, E:

The value of $Q$ is ${\text{4×10}}^{\text{1}}$.  It is greater than $\text{1}$.  Hence, $Q\text{ = K}$.

As the value of $\text{K}$ is increasing from left to right, the reaction progresses towards right.

(d)

Interpretation Introduction

Interpretation:

The given reaction is

  ${\text{X}}_{\text{2(g)}}{\text{+ Y}}_{\text{2(g)}}\rightleftharpoons {\text{2XY}}_{\text{(g)}}$

The value of $\text{K}$ has to be calculated at this temperature.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by $Q$.  For a reversible reaction of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ to NO}}_{\text{2}}$, the reaction quotient is represented as

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}\left(\text{g}\right)}\rightleftharpoons {\text{2NO}}_{\text{2(g)}}\\ \\ Q\text{ = }\frac{{\left({\text{NO}}_{\text{2}}\right)}^{\text{2}}}{\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)}\end{array}$

At equilibrium, $Q\text{ = K}$.

Explanation of Solution

As $Q\text{ = K}$,

$Q{\text{ = K = 4×10}}^{\text{1}}$.

(e)

Interpretation Introduction

Interpretation:

The given filmstrip contains five molecular scenes of a gaseous mixture as it reaches equilibrium over time.  In the filmstrip, $\text{X}$ is purple and $\text{Y}$ is orange.

Figure 1

If ${\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{< 0}$, the scene that represents the mixture at a higher temperature has to be explained.

Concept Introduction:

Change in equilibrium due to temperature changes:

If the temperature is increased for the system, the equilibrium shifts away from the heat because of the reaction needs extra heat to use.

If the temperature is decreased for the system, the equilibrium shifts towards the heat because the heat needs to be produced to make up for the loss.

Explanation of Solution

Among the given five scenes, the scene that represents the mixture at a higher temperature is scene B.  At higher temperatures, reaction shifts towards left forming more ${\text{X}}_{\text{2}}{\text{ and Y}}_{\text{2}}$.

(f)

Interpretation Introduction

Interpretation:

The given filmstrip contains five molecular scenes of a gaseous mixture as it reaches equilibrium over time.  In the filmstrip, $\text{X}$ is purple and $\text{Y}$ is orange.

Figure 1

The scene that represents the mixture at a higher pressure (lower volume) has to be explained.

Concept Introduction:

Change in equilibrium due to pressure changes:

On increase in the system pressure, the equilibrium shifts towards fewer moles of gas, because, for gases,

  $\text{increase in pressure = decrease in volume}\text{.}$

On decrease in the system pressure, the equilibrium shifts towards more moles of gas, because, for gases,

  $\text{decrease in pressure = increase in volume}\text{.}$

Explanation of Solution

The given reaction is

  ${\text{X}}_{\text{2(g)}}{\text{+ Y}}_{\text{2(g)}}\rightleftharpoons {\text{2XY}}_{\text{(g)}}$

The gaseous molecules in reactants are $\text{2 mols}$ and in products are $\text{2 mols}$.  As the moles is equal, the increase in pressure or decrease in volume does not affect the equilibrium.