Chapter 19, Problem 104CP

(a)

Interpretation Introduction

Interpretation: It is given that, nitrogen reacts with hydrogen gas (in a container) to produce ammonia gas. The volume of this mixture and total pressure is given. The partial pressure and mole fraction of ammonia in a container during completion of reaction is to be calculated. The volume of container during the completion of reaction is to be calculated.

Concept introduction: The volume is calculated using ideal gas law,

$PV=nRT$

The mole fraction of ammonia is calculated using the formula,

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}$

To determine: The partial pressure of ammonia in a container during completion of reaction.

Explanation of Solution

Given

Total pressure is $1.0\text{atm}$.

The reaction of formation of ammonia is,

${\text{N}}_2\left(g\right)+3{\text{H}}_2\left(g\right)\stackrel{}{\to }{\text{2NH}}_3\left(g\right)$

It is clear from the above equation that, one mole of nitrogen reacts with three moles of hydrogen to yield two moles of ammonia. In this reaction, hydrogen gas is limiting whereas nitrogen gas is in excess. Hence, the chemical equivalence is,

$\begin{array}{c}1\text{mol}{\text{N}}_2\cong 3\text{mol}{\text{H}}_2\\ 3\text{mol}{\text{H}}_2\cong 2\text{mol}{\text{NH}}_3\end{array}$

In the given container, ammonia is formed due to combination of six moles of nitrogen gas and six moles of hydrogen gas. In this reaction, four moles of ammonia is produced due to consumption of two moles of nitrogen gas.

The number of moles of nitrogen gas consumed to form ammonia is calculated as,

$\text{Moles}\text{of}{\text{N}}_{\text{2}}\text{required}=\text{Total}\text{moles}\text{of}\text{hydrogen}\times \frac{\text{Moles}\text{of}\text{nitrogen}}{\text{Moles}\text{of}\text{hydrogen}}$

Substitute the values of total moles of hydrogen, moles of nitrogen and hydrogen during ammonia formation in the above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{N}}_{\text{2}}\text{required}=\text{6}\text{mol}{\text{H}}_2\times \frac{\text{1}\text{mol}{\text{N}}_2}{\text{3}\text{mol}{\text{H}}_2}\\ =\underset{\_}{2mol}\end{array}$

Remaining moles of nitrogen gas is calculated by subtracting total moles of nitrogen gas with moles of nitrogen consumed. Hence, remaining moles of nitrogen gas is,

$6\text{mol}{\text{N}}_2-2\text{mol}{\text{N}}_2=\underset{\_}{4mol}$

The number of moles of ammonia produced is calculated as,

$\text{Moles}\text{of}{\text{NH}}_{\text{3}}\text{produced}=\text{Total}\text{moles}\text{of}\text{hydrogen}\times \frac{\text{Moles}\text{of}\text{ammonia}}{\text{Moles}\text{of}\text{hydrogen}}$

Substitute the values of total moles of hydrogen, moles of ammonia produced and hydrogen in the above equation.

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_{\text{3}}\text{produced}=\text{Total}\text{moles}\text{of}\text{hydrogen}\times \frac{\text{Moles}\text{of}\text{ammonia}}{\text{Moles}\text{of}\text{hydrogen}}\\ =\text{6}\text{mol}{\text{H}}_2\times \frac{\text{2}\text{mol}{\text{NH}}_{\text{3}}}{\text{3}\text{mol}{\text{H}}_2}\\ =\underset{\_}{4mol}\end{array}$

The mole fraction of ammonia is $\underset{\_}{0.5}$.

Remaining moles of nitrogen gas is $0.4\text{mol}$.

Moles of ammonia gas is $0.4\text{mol}$.

Formula

The mole fraction of ammonia is calculated using the formula,

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}$

Substitute the value of moles of ammonia and nitrogen gas in the above expression.

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}\\ =\frac{4}{4+4}\\ =\underset{\_}{0.5}\end{array}$

The partial pressure of ammonia is $\underset{\_}{0.5atm}$.

Total pressure is $1.0\text{atm}$.

Mole fraction of ammonia gas is $0.5$.

Formula

The partial pressure of ammonia is calculated using the formula,

$\text{Partial}\text{pressure}=\text{Mole}\text{fraction}\times \text{Total}\text{pressure}$

Substitute the values of mole fraction of ammonia and total pressure in the above equation.

$\begin{array}{c}\text{Partial}\text{pressure}=\text{Mole}\text{fraction}\times \text{Total}\text{pressure}\\ =\text{0}\text{.5}\times \text{1}\text{.0}\text{atm}\\ =\underset{\_}{0.5atm}\end{array}$

The number of moles of ${\text{N}}_2$ consumed is $\underset{\_}{2mol}$. Remaining moles of ${\text{N}}_2$ after completion of reaction is $\underset{\_}{4mol}$. Moles of ${\text{NH}}_3$ produced is $\underset{\_}{4mol}$.

(b)

Interpretation Introduction

Interpretation: It is given that, nitrogen reacts with hydrogen gas (in a container) to produce ammonia gas. The volume of this mixture and total pressure is given. The partial pressure and mole fraction of ammonia in a container during completion of reaction is to be calculated. The volume of container during the completion of reaction is to be calculated.

Concept introduction: The volume is calculated using ideal gas law,

$PV=nRT$

The mole fraction of ammonia is calculated using the formula,

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}$

To determine: The mole fraction of ammonia in a container during completion of reaction.

Explanation of Solution

The mole fraction of ammonia is $\underset{\_}{0.5}$.

Remaining moles of nitrogen gas is $0.4\text{mol}$.

Moles of ammonia gas is $0.4\text{mol}$.

Formula

The mole fraction of ammonia is calculated using the formula,

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}$

Substitute the value of moles of ammonia and nitrogen gas in the above expression.

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}\\ =\frac{4}{4+4}\\ =\underset{\_}{0.5}\end{array}$

(c)

Interpretation Introduction

Interpretation: It is given that, nitrogen reacts with hydrogen gas (in a container) to produce ammonia gas. The volume of this mixture and total pressure is given. The partial pressure and mole fraction of ammonia in a container during completion of reaction is to be calculated. The volume of container during the completion of reaction is to be calculated.

Concept introduction: The volume is calculated using ideal gas law,

$PV=nRT$

The mole fraction of ammonia is calculated using the formula,

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{Moles }{\text{of NH}}_3}{{\text{Moles of N}}_2+{\text{Moles of NH}}_3}$

To determine: The volume of container during the completion of reaction.

Explanation of Solution

Given

Total pressure is $1.0\text{atm}$.

Volume is $15.0\text{L}$.

It is given that pressure is constant. The number of moles decreases from its initial value $\left(12\right)$ to final value $\left(8\right)$. As a result, volume will decrease.

At initial moles, the temperature is calculated as,

$\begin{array}{c}PV=nRT\\ T=\frac{PV}{nR}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,V,R$ and $n$ in the above equation.

$\begin{array}{c}T=\frac{PV}{nR}\\ =\frac{\left(\text{1}\text{.00}\text{atm}\right)\left(\text{15}\text{.0}\text{L}\right)}{\text{12}\text{mol}(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})}\\ =\underset{\_}{15.2K}\end{array}$

Final volume during the completion of reaction is $\underset{\_}{9.98L}$.

Total pressure is $1.0\text{atm}$.

Volume is $15.0\text{L}$.

Final moles are $8\text{mol}$.

Temperature is $15.2\text{K}$.

Formula

For final moles, the final volume is calculated as,

$\begin{array}{c}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,T,R$ and $n$ in the above equation.

$\begin{array}{c}V=\frac{nRT}{P}\\ =\frac{(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})\times \text{8}\text{mol}\times \text{15}\text{.2}\text{K}}{\text{1}\text{.00}\text{atm}}\\ =\underset{\_}{9.98L}\end{array}$