Chapter 12, Problem 5ALQ

Consider the reaction $\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$ in a 1.0-L rigid flask. Answer the following questions for each situation (a–d):

i. Estimate a range (as, small as possible) for the requested substance. For example, [A] could be between 95 M and 100 M.

ii. Explain how you decided on the limits for the estimated range.

iii. Indicate what other information would enable you to narrow your estimated range.

iv. Compare the estimated concentrations for a through d. and explain any differences.

a. If at equilibrium [A]= 1 M. and then 1 mole of C is added, estimate the value for [A] once equilibrium is reestablished.

b. If at equilibrium [B] = 1 M, and then 1 mole of C is added, estimate the value for [B] once equilibrium is reestablished.

c. If at equilibrium [C] = 1 M, and then 1 mole of C is added, estimate the value for [C] once equilibrium is reestablished.

d. If at equilibrium [D] = 1 M, and then 1 mole of C is added, estimate the value for [D] once equilibrium is reestablished.

Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

To determine: The value of $\left[\text{A}\right]$ once equilibrium is reestablished if at equilibrium $\left[\text{A}\right]=\text{1 M}$ and then $1$ mole of $\text{C}$ is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

The equilibrium concentration of $\left[\text{A}\right]=\left[\text{A}+\frac{1}{\text{C}}\right]$

Explanation of Solution

Explanation

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The equilibrium constant $K$ for the above reaction is,

$K=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$

It is given that at equilibrium,   $\left[\text{A}\right]\text{=1 M}$ and $1$ mole of $\text{C}$ is added. Let the change on concentration is assumed to be $x$. Therefore, the equilibrium concentrations become,

$\left[\text{A}\right]\text{=1 M+}x$

$\left[\text{C}\right]=\left[\text{C+1}\right]$

The new equilibrium constant $\left(K^{\text{'}}\right)$ for the reaction is,

$K^{\text{'}}=\frac{\left[\text{C+1}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$

Since the equilibrium is reestablished. Therefore, the new equilibrium constant $\left(K^{\text{'}}\right)$ will be equal to $K$.

$\begin{array}{c}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}=\frac{\left[\text{C+1}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}\\ =\frac{\left[\text{C+1}\right]\left[\text{D}\right]}{\left[1+x\right]{\left[\text{B}\right]}^2}\\ \text{C}=\frac{\text{C}+1}{1+x}\\ x=\frac{1}{\text{C}}\end{array}$

Substitute the value of $x$ in the concentration of $\text{A}$. Therefore, the equilibrium concentration of $\left[\text{A}\right]=\left[\text{A}+\frac{1}{\text{C}}\right]$.

(a)(ii)

To determine: An explanation for deciding the limits for the estimated range.

Answer

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

In each of the given case, one mole of product $\text{C}$ is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants $\text{B}$ and $\text{A}$ will be higher than products $\text{C}$ and $\text{D}$.

Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

To determine: The value of $\left[\text{B}\right]$ once equilibrium is reestablished if at equilibrium $\left[\text{B}\right]=\text{1 M}$ and then $1$ mole of $\text{C}$ is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

The equilibrium concentration of $\left[\text{B}\right]=\sqrt{\text{C+1}}-1$.

Explanation of Solution

Explanation

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The equilibrium constant $K$ for the above reaction is,

$K=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$

It is given that at equilibrium, $\left[\text{B}\right]\text{=1 M}$ and $1$ mole of $\text{C}$ is added. Let the change on concentration is assumed to be $x$. Therefore, the equilibrium concentrations become,

$\left[\text{B}\right]\text{=}{\left(\text{1 M+}x\right)}^2$

$\left[\text{C}\right]=\left[\text{C+1}\right]$

The new equilibrium constant $\left(K^{\text{'}}\right)$ for the reaction is,

$K^{\text{'}}=\frac{\left[\text{C+1}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$

Since the equilibrium is reestablished. Therefore, the new equilibrium constant $\left(K^{\text{'}}\right)$ will be equal to $K$.

$\begin{array}{c}\frac{\left[\text{C+1}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[1+x\right]}^2}=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}\\ \frac{\text{C}+1}{{\left(1+x\right)}^2}=\left[\text{C}\right]\\ \text{B}=\sqrt{\text{C+1}}-1\end{array}$

Therefore, the equilibrium concentration of $\left[\text{B}\right]=\sqrt{\text{C+1}}-1$.

(ii)

To determine: An explanation for deciding the limits for the estimated range.

Solution

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

In each of the given case, one mole of product $\text{C}$ is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants $\text{B}$ and $\text{A}$ will be higher than products $\text{C}$ and $\text{D}$.

Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

To determine: The value of $\left[\text{C}\right]$ once equilibrium is reestablished if at equilibrium $\left[\text{C}\right]=\text{1 M}$ and then $1$ mole of $\text{C}$ is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

There will be a change in concentration of $\text{C}$.

Explanation of Solution

Explanation

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The equilibrium constant $K$ for the above reaction is,

$K=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$

It is given that at equilibrium, $\left[\text{C}\right]\text{=1 M}$ and $1$ mole of $\text{C}$ is added. Let the change on concentration is assumed to be $x$. Therefore, the equilibrium concentrations become,

$\left[\text{C}\right]=\left[\text{C+1}\right]$

Therefore, the equilibrium concentration of $\left[\text{C}\right]=\left[\text{C+1}\right]$.

(ii)

To determine: An explanation for deciding the limits for the estimated range.

Answer

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

In each of the given case, one mole of product $\text{C}$ is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants $\text{B}$ and $\text{A}$ will be higher than products $\text{C}$ and $\text{D}$.

Interpretation Introduction

Interpretation: Answers for the questions for each of the given situations are to be stated.

Concept introduction: Chemical equilibrium is a state of a system in which the rate of forward and backward reactions is equal. It is affected by various factors such as concentration of reactants or products, temperature and pressure.

(i)

To determine: The value of $\left[\text{D}\right]$ once equilibrium is reestablished if at equilibrium $\left[\text{D}\right]=\text{1 M}$ and then $1$ mole of $\text{C}$ is added and estimate a range for the requested substance.

Answer to Problem 5ALQ

Answer

There will be a change in concentration of $\text{D}$.

Explanation of Solution

Explanation

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

It is given that at equilibrium, $\left[\text{D}\right]=\text{1 M}$ and $1$ mole of $\text{C}$ is added. Let the change on concentration is assumed to be $x$. Therefore, the equilibrium concentrations become,

$\left[\text{D}\right]=\text{1 M+}x$

$\left[\text{C}\right]=\left[\text{C+1}\right]$

The new equilibrium constant $\left(K^{\text{'}}\right)$ for the reaction is,

$K^{\text{'}}=\frac{\left[\text{C+1}\right]\left[1+x\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$

Since the equilibrium is reestablished. Therefore, the new equilibrium constant $\left(K^{\text{'}}\right)$ will be equal to $K$.

$\begin{array}{c}\frac{\left[\text{C+1}\right]\left[1+x\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}\\ \left(\text{C}+1\right)\left(1+x\right)=\text{C}\times \left[\text{D}\right]\\ \left[\text{D}\right]=\frac{\text{C}}{\left(\text{C}+1\right)\left(1+x\right)}\end{array}$

Therefore, the equilibrium concentration of $\left[\text{D}\right]=\frac{\text{C}}{\left(\text{C}+1\right)\left(1+x\right)}$.

(ii)

To determine: An explanation for deciding the limits for the estimated range.

Answer

The limits were decided for the estimated range.

The equilibrium constant expression is important in deciding the limits for the estimated range. It allows us to predict several important features of the reaction: the tendency of the reaction to occur but it does not tell about the speed of the reaction.

(iii)

To determine: The other information that enable to narrow the estimated range.

Answer

The other information involves temperature and pressure.

The equilibrium constant changes with temperature. If energy is added to a system in equilibrium, the position of equilibrium will be in the direction that consumes energy. Therefore, the equilibrium will shift in reverse direction. When pressure of a system in equilibrium is increased, the equilibrium will shift toward the side having fewer moles of gas.

(iv)

To determine: The comparison of estimated concentrations for a through d and explanation for the differences.

Answer

The comparison of estimated concentrations for a through d is done.

The given reaction is,

$\text{A}\left(g\right)+2\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

In each of the given case, one mole of product $\text{C}$ is added. Addition of product will shift the equilibrium in reverse direction. According to Le Chatelier principle, when product is added to a system in equilibrium, the equilibrium will shift in reverse direction to balance the effect. Thus concentration of reactants $\text{B}$ and $\text{A}$ will be higher than products $\text{C}$ and $\text{D}$.