Chapter 6, Problem 83AE

(a)

Interpretation Introduction

Interpretation:Partial pressure of gases ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ after equilibrium is to be determined.

Concept introduction:Partial pressure of gas is defined as the product of total pressure with mole fraction of gas. Equilibrium constant for partial pressure is denoted by $K_{\text{p}}$ and is equal to ratio of pressure of product to pressure of reactant. Consider an equilibrium equation as follows:

  $\text{A}\rightleftharpoons \text{B}$

The expression used to calculate $K_{\text{p}}$ is as follows:

  $K_{\text{p}}=\frac{P_{\text{B}}}{P_{\text{A}}}$

Where,

• $K_{\text{p}}$ is equilibrium constant for partial pressure.
• $P_{\text{B}}$ is partial pressure of B (product).
• $P_{\text{A}}$ is partial pressure of A (reactant).

Answer to Problem 83AE

Value of $P_{{\text{CO}}_2}$ and $P_{{\text{H}}_{\text{2}}\text{O}}$ is $0.5\text{ atm}$ .

Explanation of Solution

Given reaction is as follows:

  $2{\text{NaHCO}}_{\text{3}}\left(s\right)\rightleftharpoons {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The ICE table for givenreaction is as follows:

  $\begin{array}{cccccccc}& 2{\text{NaHCO}}_{\text{3}}\left(s\right)& \rightleftharpoons & {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)& +& C{O}_2\left(g\right)& +& {\text{H}}_{\text{2}}\text{O}\left(g\right)\\ \text{Initial}& -& & -& & 0& & 0\\ \text{Change}& -& & -& & x& & x\\ \text{Equilibrium}& -& & -& & x& & x\end{array}$

Solid terms are not considered in ICE table as they are not affected by pressure. Therefore, partial pressure of ${\text{CO}}_2\left(g\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is equal. Consider $P_{{\text{CO}}_2}$ and $P_{{\text{H}}_{\text{2}}\text{O}}$ to be $x$ .

The expression for equilibrium constant of partial pressure, $K_{\text{p}}$ is as follows:

  $K_{\text{p}}=\left(P_{{\text{CO}}_2}\right)\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)$

Rearrange above equation for $P_{{\text{CO}}_2}$ .

  $P_{{\text{CO}}_2}=\frac{K_{\text{p}}}{P_{{\text{H}}_{\text{2}}\text{O}}}$

Value of $P_{{\text{H}}_{\text{2}}\text{O}}$ is $x$ .

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

Value of $K_{\text{p}}$ is $0.25$ .

Substitute values in above equation.

  $\begin{array}{c}{P}_{{\text{CO}}_2}=\frac{K_{\text{p}}}{P_{{\text{H}}_{\text{2}}\text{O}}}\\ x=\frac{0.25}{x}\\ x^2=0.25\\ x=0.5\end{array}$

Thus, value of $P_{{\text{CO}}_2}$ and $P_{{\text{H}}_{\text{2}}\text{O}}$ is $0.5\text{ atm}$ .

(b)

Interpretation Introduction

Interpretation: Mass of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_2{\text{CO}}_{\text{3}}$ at equilibrium is to be calculated.

Concept introduction:The ideal gas equation can be expressed as follows,

  $PV=nRT$

Where,

• $P$ is the pressure of gas.
• $V$ is the volume of gas
• $T$ is the temperature of gas.
• $n$ is the mole of gas.
• $R$ is the gas constant.

Mole is defined as the ratio of mass of a substance to its molar mass. It is expressed as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

Answer to Problem 83AE

Mass of ${\text{NaHCO}}_{\text{3}}$ is $\text{7}\text{.5 g}$ and ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is $1.6\text{ g}$ .

Explanation of Solution

Given reaction is as follows:

  $2{\text{NaHCO}}_{\text{3}}\left(s\right)\rightleftharpoons {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Number of moles of ${\text{CO}}_2$ from ideal gas equation can be expressed as follows,

  $PV=nRT$

Rearrange above equation for $n$ .

  $n=\frac{RT}{PV}$

Where,

• $P$ is the pressure of ${\text{CO}}_2$ .
• $V$ is the volume of container.
• $T$ is the temperature of gas.
• $n$ is the mole of gas.
• $R$ is the gas constant.

Value of $P$ is $0.5\text{ atm}$ .

Value of $V$ is $1\text{ L}$ .

Value of $R$ is $0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}$ .

Value of $T$ is $398\text{ K}$ .

Substitute values in above equation.

  $\begin{array}{c}n=\frac{RT}{PV}\\ =\frac{\left(0.0821\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(398\text{ K}\right)}{\left(0.5\text{ atm}\right)\left(1\text{ L}\right)}\\ =0.015\text{ mol}\end{array}$

Reaction of two moles ${\text{NaHCO}}_{\text{3}}$ produces one mole of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ , one mole of ${\text{CO}}_2$ and one mole of ${\text{H}}_{\text{2}}\text{O}$ . Therefore $0.015\text{ mol}$ of ${\text{CO}}_2$ produces $0.030\text{ mol}$ of ${\text{NaHCO}}_{\text{3}}$ and $0.015\text{ mol}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ . Hence, formula used for calculation of mass of ${\text{NaHCO}}_{\text{3}}$ is as follows:

  ${\text{Mole of NaHCO}}_{\text{3}}=\frac{{\text{mass of NaHCO}}_{\text{3}}}{{\text{molar mass of NaHCO}}_{\text{3}}}$

Rearrange above equation for mass of ${\text{NaHCO}}_{\text{3}}$ .

  ${\text{Mass of NaHCO}}_{\text{3}}=\left({\text{Mole of NaHCO}}_{\text{3}}\right)\left({\text{molar mass of NaHCO}}_{\text{3}}\right)$

Value of moles of ${\text{NaHCO}}_{\text{3}}$ is $0.030\text{ mol}$ .

Molar mass of ${\text{NaHCO}}_{\text{3}}$ is $84.00\text{ g/mol}$ .

Substitute values in above equation.

  $\begin{array}{c}{\text{Mass of NaHCO}}_{\text{3}}=\left({\text{Mole of NaHCO}}_{\text{3}}\right)\left({\text{molar mass of NaHCO}}_{\text{3}}\right)\\ =\left(0.030\text{ mol}\right)\left(84.00\text{ g/mol}\right)\\ =2.52\text{ g}\end{array}$

Expression for calculation of mass of ${\text{NaHCO}}_{\text{3}}$ at equilibrium is as follows:

  ${\text{Mass of NaHCO}}_{\text{3}}\text{ at equilibrium}={\text{mass of initial NaHCO}}_{\text{3}}-{\text{mass of NaHCO}}_{\text{3}}$

Value of mass of initial ${\text{NaHCO}}_{\text{3}}$ is $10\text{ g}$ .

Value of mass of ${\text{NaHCO}}_{\text{3}}$ is $2.52\text{ g}$ .

Substitute values in above equation.

  $\begin{array}{c}{\text{Mass of NaHCO}}_{\text{3}}\text{ at equilibrium}={\text{mass of initial NaHCO}}_{\text{3}}-{\text{mass of NaHCO}}_{\text{3}}\\ =\left(10-2.52\right)\text{ g}\\ =\text{7}\text{.5 g}\end{array}$

Formula used for calculation of mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

  ${\text{Mole of Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{{\text{mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{{\text{molar mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}}$

Rearrange above equation for mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ .

  ${\text{Mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\left({\text{Mole of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\left({\text{molar mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$

Value of moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $0.015\text{ mol}$ .

Molar mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $105.98\text{ g/mol}$ .

Substitute values in above equation.

  $\begin{array}{c}{\text{Mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\left({\text{Mole of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\left({\text{molar mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\\ =\left(0.015\text{ mol}\right)\left(105.98\text{ g/mol}\right)\\ =1.6\text{ g}\end{array}$

Hence, mass of ${\text{NaHCO}}_{\text{3}}$ is $\text{7}\text{.5 g}$ and ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is $1.6\text{ g}$ .

(c)

Interpretation Introduction

Interpretation:Minimum volume required by container to decompose ${\text{NaHCO}}_{\text{3}}$ is to be calculated.

Concept introduction: Mole is defined as the ratio of mass of a substance to its molar mass. It is expressed as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

Answer to Problem 83AE

Minimum volume required by container to decompose ${\text{NaHCO}}_{\text{3}}$ is $\text{3}\text{.9 L}$ .

Explanation of Solution

Formula used for calculation of initial moles of ${\text{NaHCO}}_{\text{3}}$ is as follows:

  ${\text{Initial mole of NaHCO}}_{\text{3}}=\frac{{\text{mass of NaHCO}}_{\text{3}}}{{\text{molar mass of NaHCO}}_{\text{3}}}$

Value of mass of ${\text{NaHCO}}_{\text{3}}$ is $10\text{ g}$ .

Molar mass of ${\text{NaHCO}}_{\text{3}}$ is $84.00\text{ g/mol}$ .

Substitute values in above equation.

  $\begin{array}{c}{\text{Initial mole of NaHCO}}_{\text{3}}=\frac{{\text{mass of NaHCO}}_{\text{3}}}{{\text{molar mass of NaHCO}}_{\text{3}}}\\ =\frac{10\text{ g}}{84\text{ g/mol}}\\ =0.119\text{ mol}\end{array}$

Given reaction is as follows:

  $2{\text{NaHCO}}_{\text{3}}\left(s\right)\rightleftharpoons {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Number of moles of ${\text{CO}}_2$ from ideal gas equation can be expressed as follows,

  $PV=nRT$

Rearrange above equation for $n$ .

  $n=\frac{RT}{PV}$

Where,

• $P$ is the pressure of ${\text{CO}}_2$ .
• $V$ is the volume of container.
• $T$ is the temperature of gas.
• $n$ is the mole of gas.
• $R$ is the gas constant.

Value of $P$ is $0.5\text{ atm}$ .

Value of $V$ is $1\text{ L}$ .

Value of $R$ is $0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}$ .

Value of $T$ is $398\text{ K}$ .

Substitute values in above equation.

  $\begin{array}{c}n=\frac{RT}{PV}\\ =\frac{\left(0.0821\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(398\text{ K}\right)}{\left(0.5\text{ atm}\right)\left(1\text{ L}\right)}\\ =0.015\text{ mol}\end{array}$

Reaction of two moles ${\text{NaHCO}}_{\text{3}}$ produces one mole of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ , one mole of ${\text{CO}}_2$ and one mole of ${\text{H}}_{\text{2}}\text{O}$ . Therefore $0.015\text{ mol}$ of ${\text{CO}}_2$ produces $0.030\text{ mol}$ of ${\text{NaHCO}}_{\text{3}}$ .

The expression used to determine the concentration of ${\text{NaHCO}}_{\text{3}}$ is as follows:

  ${\text{Concentration of NaHCO}}_{\text{3}}=\frac{{\text{moles of NaHCO}}_{\text{3}}}{\text{volume of solution}\left(\text{L}\right)}$

Value of moles of ${\text{NaHCO}}_{\text{3}}$ is $0.030\text{ mol}$ .

Value of volume of solutionis $1\text{ L}$ .

Substitute values in above equation.

  $\begin{array}{c}{\text{Concentration of NaHCO}}_{\text{3}}=\frac{{\text{moles of NaHCO}}_{\text{3}}}{\text{volume of solution}\left(\text{L}\right)}\\ =\frac{0.030\text{ mol}}{1\text{ L}}\\ =0.030\text{ mol/L}\end{array}$

Hence, moles of ${\text{NaHCO}}_{\text{3}}$ per litreis $0.030\text{ mol/L}$ .

The expression used for calculation of volume required to decompose ${\text{NaHCO}}_{\text{3}}$ is as follows:

  ${\text{Volume of NaHCO}}_3=\frac{{\text{initial moles of NaHCO}}_{\text{3}}}{{\text{moles of NaHCO}}_3\text{ per L}}$

Value of initial moles of ${\text{NaHCO}}_{\text{3}}$ is $0.119\text{ mol}$ .

Value of moles of ${\text{NaHCO}}_{\text{3}}$ per litreis $0.030\text{ mol/L}$ .

Substitute values in above equation.

  $\begin{array}{c}{\text{Volume of NaHCO}}_3=\frac{0.119\text{ mol}}{0.030\text{ mol/L}}\\ =\text{3}\text{.9 L}\end{array}$