Chapter 6, Problem 39E

a)

Interpretation Introduction

Interpretation: Whether initial amount of $\text{CaO}$ increase, decrease or remain same if $655\text{ g}$ of ${\text{CaCO}}_{\text{3}}$ , $95.0\text{ g}$ of $\text{CaO}$ and $58.4\text{ g}$ of ${\text{CO}}_{\text{2}}$ are present in reaction chamber at $900°\text{C}$ should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Reaction quotient is determined with relative amounts of reactants as well as products. It is denoted by $Q_{\text{p}}$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q_{\text{p}}$ and $K_{\text{p}}$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q_{\text{p}}$ and $K_{\text{p}}$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q_{\text{p}}$ is greater than $K_{\text{p}}$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q_{\text{p}}$ is less than $K_{\text{p}}$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Expression for moles of ${\text{CO}}_{\text{2}}$ is as follows:

  $n=\frac{m}{M}$

Where,

• $n$ is number of moles of ${\text{CO}}_{\text{2}}$ .
• $m$ is mass of .
• $M$ is molar mass of ${\text{CO}}_{\text{2}}$ .

Value of $m$ is $58.4\text{ g}$ .

Value of $M$ is $44.01\text{ g/mol}$ .

Substitute the values in above equation,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{58.4\text{ g}}{44.01\text{ g/mol}}\\ =1.327\text{ mol}\end{array}$

Expression for ideal gas equation for ${\text{CO}}_{\text{2}}$ is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of ${\text{CO}}_{\text{2}}$ .
• $V$ is volume of ${\text{CO}}_{\text{2}}$ .
• $n$ is amount or moles of ${\text{CO}}_{\text{2}}$ .
• $R$ is universal gas constant.
• $T$ is absolute temperature of ${\text{CO}}_{\text{2}}$ .

Rearrange above equation for $P$ .

  $P=\frac{nRT}{V}$

Temperature of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}T=\left(900+273\right)\text{ K}\\ =1173\text{ K}\end{array}$

Value of $V$ is $\text{50}\text{.0 L}$ .

Value of $R$ is $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $T$ is $1173\text{ K}$ .

Value of $n$ is $1.327\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{P}_{{\text{CO}}_2}=\frac{nRT}{V}\\ =\frac{\left(1.327\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(1173\text{ K}\right)}{50.0\text{ L}}\\ =2.555\text{ atm}\end{array}$

Expression for $Q_{\text{p}}$ is as follows:

  $Q_{\text{p}}=P_{{\text{CO}}_2}$

Where,

• $Q_{\text{p}}$ is reaction quotient.
• $P_{{\text{CO}}_2}$ is pressure of ${\text{CO}}_2$ .

Value of $P_{{\text{CO}}_2}$ is $2.555\text{ atm}$ .

Substitute the values in above equation.

  $\begin{array}{c}{Q}_{\text{p}}=P_{{\text{CO}}_2}\\ =2.555\end{array}$

But value of $K$ is 1.04. Since value of $Q$ is greater than that of $K$ , shift in left direction is observed. This means consumption of $\text{CaO}$ will be more. Therefore initial amount of $\text{CaO}$ decreases as system moves towards equilibrium.

b)

Interpretation Introduction

Interpretation: Whether initial amount of $\text{CaO}$ increase, decrease or remain same if $780\text{ g}$ of ${\text{CaCO}}_{\text{3}}$ , $1.00\text{ g}$ of $\text{CaO}$ and $23.76\text{ g}$ of ${\text{CO}}_{\text{2}}$ are present in reaction chamber at $900°\text{C}$ should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Reaction quotient is determined with relative amounts of reactants as well as products. It is denoted by $Q_{\text{p}}$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q_{\text{p}}$ and $K_{\text{p}}$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q_{\text{p}}$ and $K_{\text{p}}$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q_{\text{p}}$ is greater than $K_{\text{p}}$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q_{\text{p}}$ is less than $K_{\text{p}}$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Expression for moles of ${\text{CO}}_{\text{2}}$ is as follows:

  $n=\frac{m}{M}$

Where,

• $n$ is number of moles of ${\text{CO}}_{\text{2}}$ .
• $m$ is mass of .
• $M$ is molar mass of ${\text{CO}}_{\text{2}}$ .

Value of $m$ is $23.76\text{ g}$ .

Value of $M$ is $44.01\text{ g/mol}$ .

Substitute the values in above equation,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{23.76\text{ g}}{44.01\text{ g/mol}}\\ =0.5399\text{ mol}\end{array}$

Expression for ideal gas equation for ${\text{CO}}_{\text{2}}$ is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of ${\text{CO}}_{\text{2}}$ .
• $V$ is volume of ${\text{CO}}_{\text{2}}$ .
• $n$ is amount or moles of ${\text{CO}}_{\text{2}}$ .
• $R$ is universal gas constant.
• $T$ is absolute temperature of ${\text{CO}}_{\text{2}}$ .

Rearrange above equation for $P$ .

  $P=\frac{nRT}{V}$

Temperature of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}T=\left(900+273\right)\text{ K}\\ =1173\text{ K}\end{array}$

Value of $V$ is $\text{50}\text{.0 L}$ .

Value of $R$ is $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $T$ is $1173\text{ K}$ .

Value of $n$ is $0.5399\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{P}_{{\text{CO}}_2}=\frac{nRT}{V}\\ =\frac{\left(0.5399\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(1173\text{ K}\right)}{50.0\text{ L}}\\ =1.04\text{ atm}\end{array}$

Expression for $Q_{\text{p}}$ is as follows:

  $Q_{\text{p}}=P_{{\text{CO}}_2}$

Where,

• $Q_{\text{p}}$ is reaction quotient.
• $P_{{\text{CO}}_2}$ is pressure of ${\text{CO}}_2$ .

Value of $P_{{\text{CO}}_2}$ is $1.04\text{ atm}$ .

Substitute the values in above equation.

  $\begin{array}{c}{Q}_{\text{p}}=P_{{\text{CO}}_2}\\ =1.04\end{array}$

But value of $K$ is 1.04. Since value of $Q$ is equal to $K$ , no shift in direction is observed. Therefore initial amount of $\text{CaO}$ remains same as system moves towards equilibrium.

c)

Interpretation Introduction

Interpretation: Whether initial amount of $\text{CaO}$ increase, decrease or remain same if $0.14\text{ g}$ of ${\text{CaCO}}_{\text{3}}$ , $5000\text{ g}$ of $\text{CaO}$ and $23.76\text{ g}$ of ${\text{CO}}_{\text{2}}$ are present in reaction chamber at $900°\text{C}$ should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Reaction quotient is determined with relative amounts of reactants as well as products. It is denoted by $Q_{\text{p}}$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q_{\text{p}}$ and $K_{\text{p}}$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q_{\text{p}}$ and $K_{\text{p}}$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q_{\text{p}}$ is greater than $K_{\text{p}}$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q_{\text{p}}$ is less than $K_{\text{p}}$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Expression for moles of ${\text{CO}}_{\text{2}}$ is as follows:

  $n=\frac{m}{M}$

Where,

• $n$ is number of moles of ${\text{CO}}_{\text{2}}$ .
• $m$ is mass of .
• $M$ is molar mass of ${\text{CO}}_{\text{2}}$ .

Value of $m$ is $23.76\text{ g}$ .

Value of $M$ is $44.01\text{ g/mol}$ .

Substitute the values in above equation,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{23.76\text{ g}}{44.01\text{ g/mol}}\\ =0.5399\text{ mol}\end{array}$

Expression for ideal gas equation for ${\text{CO}}_{\text{2}}$ is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of ${\text{CO}}_{\text{2}}$ .
• $V$ is volume of ${\text{CO}}_{\text{2}}$ .
• $n$ is amount or moles of ${\text{CO}}_{\text{2}}$ .
• $R$ is universal gas constant.
• $T$ is absolute temperature of ${\text{CO}}_{\text{2}}$ .

Rearrange above equation for $P$ .

  $P=\frac{nRT}{V}$

Temperature of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}T=\left(900+273\right)\text{ K}\\ =1173\text{ K}\end{array}$

Value of $V$ is $\text{50}\text{.0 L}$ .

Value of $R$ is $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $T$ is $1173\text{ K}$ .

Value of $n$ is $0.5399\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{P}_{{\text{CO}}_2}=\frac{nRT}{V}\\ =\frac{\left(0.5399\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(1173\text{ K}\right)}{50.0\text{ L}}\\ =1.04\text{ atm}\end{array}$

Expression for $Q_{\text{p}}$ is as follows:

  $Q_{\text{p}}=P_{{\text{CO}}_2}$

Where,

• $Q_{\text{p}}$ is reaction quotient.
• $P_{{\text{CO}}_2}$ is pressure of ${\text{CO}}_2$ .

Value of $P_{{\text{CO}}_2}$ is $1.04\text{ atm}$ .

Substitute the values in above equation.

  $\begin{array}{c}{Q}_{\text{p}}=P_{{\text{CO}}_2}\\ =1.04\end{array}$

But value of $K$ is 1.04. Since value of $Q$ is equal to $K$ , no shift in direction is observed. Therefore initial amount of $\text{CaO}$ remains same as system moves towards equilibrium.

d)

Interpretation Introduction

Interpretation: Whether initial amount of $\text{CaO}$ increase, decrease or remain same if $715\text{ g}$ of ${\text{CaCO}}_{\text{3}}$ , $813\text{ g}$ of $\text{CaO}$ and $4.82\text{ g}$ of ${\text{CO}}_{\text{2}}$ are present in reaction chamber at $900°\text{C}$ should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Reaction quotient is determined with relative amounts of reactants as well as products. It is denoted by $Q_{\text{p}}$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q_{\text{p}}$ and $K_{\text{p}}$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q_{\text{p}}$ and $K_{\text{p}}$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q_{\text{p}}$ is greater than $K_{\text{p}}$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q_{\text{p}}$ is less than $K_{\text{p}}$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Expression for moles of ${\text{CO}}_{\text{2}}$ is as follows:

  $n=\frac{m}{M}$

Where,

• $n$ is number of moles of ${\text{CO}}_{\text{2}}$ .
• $m$ is mass of .
• $M$ is molar mass of ${\text{CO}}_{\text{2}}$ .

Value of $m$ is $4.82\text{ g}$ .

Value of $M$ is $44.01\text{ g/mol}$ .

Substitute the values in above equation,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{4.82\text{ g}}{44.01\text{ g/mol}}\\ =0.1095\text{ mol}\end{array}$

Expression for ideal gas equation for ${\text{CO}}_{\text{2}}$ is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of ${\text{CO}}_{\text{2}}$ .
• $V$ is volume of ${\text{CO}}_{\text{2}}$ .
• $n$ is amount or moles of ${\text{CO}}_{\text{2}}$ .
• $R$ is universal gas constant.
• $T$ is absolute temperature of ${\text{CO}}_{\text{2}}$ .

Rearrange above equation for $P$ .

  $P=\frac{nRT}{V}$

Temperature of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}T=\left(900+273\right)\text{ K}\\ =1173\text{ K}\end{array}$

Value of $V$ is $\text{50}\text{.0 L}$ .

Value of $R$ is $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Value of $T$ is $1173\text{ K}$ .

Value of $n$ is $0.1095\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{P}_{{\text{CO}}_2}=\frac{nRT}{V}\\ =\frac{\left(0.1095\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(1173\text{ K}\right)}{50.0\text{ L}}\\ =0.2108\text{ atm}\end{array}$

Expression for $Q_{\text{p}}$ is as follows:

  $Q_{\text{p}}=P_{{\text{CO}}_2}$

Where,

• $Q_{\text{p}}$ is reaction quotient.
• $P_{{\text{CO}}_2}$ is pressure of ${\text{CO}}_2$ .

Value of $P_{{\text{CO}}_2}$ is $0.2108\text{ atm}$ .

Substitute the values in above equation.

  $\begin{array}{c}{Q}_{\text{p}}=P_{{\text{CO}}_2}\\ =0.2108\end{array}$

But value of $K$ is 1.04. Since value of $Q$ is less than that of $K$ , shift in right direction is observed. Therefore initial amount of $\text{CaO}$ increases as system moves towards equilibrium.