Chapter 13, Problem 116CP

At 125°C, K_P = 0.25 for the reaction

$2{\text{NaHCO}}_3\left(s\right)\rightleftharpoons {\text{Na}}_2{\text{CO}}_3\left(s\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

A 1.00-L flask containing 10.0 g NaHCO₃ is evacuated and heated to 125°C.

a. Calculate the par1ial pressures of CO₂ and H₂O after equilibrium is established.

b. Calculate the masses of NaHCO₃ and Na₂CO₃ present at equilibrium.

c. Calculate the minimum container volume necessary for all of the NaHCO₃ to decompose.

Interpretation Introduction

Interpretation: After equilibrium the partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_2\text{O}$ is to be calculated. At equilibrium the masses of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$  is to be calculated. For all the decomposition of ${\text{NaHCO}}_{\text{3}}$ the minimum container volume is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

To determine: The partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_2\text{O}$

Answer to Problem 116CP

Answer

1. a) The partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_2\text{O}$ is $\underset{\_}{0.50atm}$.

Explanation of Solution

Explanation

The partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_2\text{O}$ is $\underset{\_}{0.50atm}$.

Given

The reaction is as,

${\text{2NaHCO}}_3\left(s\right)\rightleftharpoons {\text{Na}}_2{\text{CO}}_3\left(s\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The equilibrium constant $K_{\text{p}}$ is $0.25$ at $125°\text{C}$.

Mass of ${\text{NaHCO}}_3$ is $10\text{g}$.

Volume of the flask is $1.00\text{L}$.

To calculate the partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ after equilibrium established, the (ICE-table) is given as,

$\begin{array}{ccccc}& 2{\text{NaHCO}}_{\text{3}}\left(s\right)\rightleftharpoons & {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+& {\text{CO}}_{\text{2}}\left(g\right)+& H_{2}O\left(g\right)\\ \text{Initial}\text{pressure}& -& -& 0.0& 0.0\\ \text{change}\text{in}\text{pressure}& -& -& +x& +x\\ \text{Equilibrium}\text{pressure}& -& -& x& x\end{array}$

Solids are not considered in ICE-chart because solids do not affect the pressure.

The equilibrium constant in terms of partial pressure for the given reaction is given by the formula.

$K_{\text{p}}=P_{{\text{CO}}_{\text{2}}}^{}{P}_{{\text{H}}_{\text{2}}\text{O}}$

Where,

• $K_{\text{p}}$ is equilibrium constant in terms of partial pressure.
• $P_{{\text{CO}}_{\text{2}}}^{}$ is the partial pressure of ${\text{CO}}_2$.
• $P_{{\text{H}}_{\text{2}}\text{O}}^{}$ is the partial pressure of ${\text{H}}_{\text{2}}\text{O}$.

Substitute the values of partial pressures from the ICE-table in the above equation.

$\begin{array}{c}{K}_{\text{p}}=P_{{\text{CO}}_{\text{2}}}^{}{P}_{{\text{H}}_{\text{2}}\text{O}}\\ 0.25=\left(x\right)\left(x\right)\\ x=\underset{\_}{0.50}\end{array}$

Therefore, the partial pressure of ${\text{CO}}_{\text{2}}$ is $\underset{\_}{0.50atm}$.

The partial pressure of ${\text{H}}_{\text{2}}\text{O}$ is $\underset{\_}{0.50atm}$.

Interpretation Introduction

Interpretation: After equilibrium the partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_2\text{O}$ is to be calculated. At equilibrium the masses of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$  is to be calculated. For all the decomposition of ${\text{NaHCO}}_{\text{3}}$ the minimum container volume is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

To determine: The masses of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ at equilibrium.

Answer to Problem 116CP

Answer

At equilibrium the masses of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ are $\underset{\_}{7.43g}$.and $\underset{\_}{1.62g}$ respectively.

Explanation of Solution

Explanation

At equilibrium the masses of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ are $\underset{\_}{7.43g}$.and $\underset{\_}{1.62g}$ respectively.

Given

The temperature is $125°\text{C}$

The relationship between degree Celsius and Kelvin is given as,

$K=°C+273$

Where,

• $K$ is the temperature in Kelvin.
• $°C$ is the temperature in degree Celsius.

Therefore, the temperature in Kelvin,

$\begin{array}{c}K=°C+273\\ =125°\text{C+273}\\ \text{=398}\text{K}\end{array}$

At equilibrium the number of moles calculated by the formula,

$n=\frac{PV}{\text{R}T}$

Where,

• $n$ is the number of moles.
• $P$ is the pressure.
• $V$ is the volume in L.
• $\text{R}$ is the gas constant.
• $T$ is the temperature in Kelvin.

Substitute the values in the above equation.

$\begin{array}{c}n=\frac{PV}{\text{R}T}\\ =\frac{\left(0.50\text{atm}\right)\left(1.00\text{L}\right)}{\left(0.08206\text{L}\cdot \text{atm}\cdot {\text{mol}}^{\text{-1}}\cdot {\text{K}}^{\text{-1}}\right)\left(398\text{K}\right)}\\ =0.0153\text{mol}\end{array}$

The mass of ${\text{NaHCO}}_{\text{3}}$ is calculated by the expression,

$\text{Mass}\text{of}{\text{NaHCO}}_{\text{3}}=M\text{ol}\text{of}{\text{NaHCO}}_{\text{3}}\times \text{Molar}\text{mass}$.

Substitute the values for $\text{NaHCO3}$,

$\begin{array}{c}0.0153\text{mol}\text{of}{\text{CO}}_{\text{2}}\times \frac{2\text{mol}\text{of}{\text{NaHCO}}_3}{\text{1}\text{mol}\text{of}{\text{CO}}_{\text{2}}}=0.0306\text{mol}\\ \text{=0}\text{.0306}\text{mol}\times 83.99\text{g}\\ \text{=2}\text{.57}\text{g}{\text{NaHCO}}_{\text{3}}\end{array}$.

Substitute the values for ${\text{Na}}_2\text{CO3}$,

$\begin{array}{c}0.0153\text{mol}\text{of}{\text{CO}}_{\text{2}}\times \frac{1\text{mol}\text{of}{\text{NaHCO}}_3}{\text{1}\text{mol}\text{of}{\text{CO}}_{\text{2}}}=0.0153\text{mol}\\ \text{=0}\text{.0306}\text{mol}\times 105.97\text{g}\\ \text{=}\underset{\_}{1.62g}{\text{Na}}_2{\text{CO}}_{\text{3}}\end{array}$

The already present amount of ${\text{NaHCO}}_{\text{3}}$ is $10\text{g}$. Therefore, the amount of ${\text{NaHCO}}_{\text{3}}$ at equilibrium is given as,

$10.00-2.57=\underset{\_}{7.43g}$.

The amount of ${\text{Na}}_2\text{CO3}$ at equilibrium is $\underset{\_}{1.62g}$.

Interpretation Introduction

Interpretation: After equilibrium the partial pressure of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_2\text{O}$ is to be calculated. At equilibrium the masses of ${\text{NaHCO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$  is to be calculated. For all the decomposition of ${\text{NaHCO}}_{\text{3}}$ the minimum container volume is to be calculated.

Concept introduction: The equilibrium constant $K_{\text{p}}$ describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure.

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

To determine: The minimum container volume to decompose the all amount of $\text{NaHCO3}$.

Answer to Problem 116CP

Answer

For all the decomposition of ${\text{NaHCO}}_{\text{3}}$ the minimum container volume is $\underset{\_}{3.89L}$.

Explanation of Solution

Explanation

For all the decomposition of ${\text{NaHCO}}_{\text{3}}$ the minimum container volume is $\underset{\_}{3.89L}$.

The volume of the container is determined by the formula,

$V_{\text{C}}=\text{Original}\text{number}\text{of}\text{moles}\times \text{moles}\text{used}\text{per}\text{L}\text{volume}$.

Where,

• $V_{\text{C}}$ is the volume of the container.

Substitute the values from the above steps.

$\begin{array}{c}{V}_{\text{C}}=\text{Original}\text{number}\text{of}\text{moles}\times \text{moles}\text{used}\text{per}\text{L}\text{volume}\\ \text{=}\frac{10\text{g}{\text{NaHCO}}_3}{83.99\text{g}{\text{NaHCO}}_3}\times \frac{1.0\text{L}}{0.0306\text{mol}{\text{NaHCO}}_3}\\ =\underset{\_}{3.89L}\end{array}$