Chapter 5, Problem 123AE

Interpretation Introduction

Interpretation:Mass of acrylonitrile formed is to be calculated.

Concept introduction:Partial pressure of a component in a gas mixture is defined as the pressure that would be exerted by that component of gas if it occupies the entire volume under the same condition.

Mole of i-th component of gas present in a gas mixture, at fixed partial pressure and temperature is calculated by the equation as follows:

  ${\text{P}}_{\text{i}}\text{V}={\text{n}}_{\text{i}}\text{RT}$

Where,

• ${\text{P}}_{\text{i}}$ is the partial pressure of i-th component gas.
• $\text{V}$ is the volume of gas container.
• ${\text{n}}_{\text{i}}$ isthe number ofmole of i-th component gas present in container.
• $\text{R}$ isthe universal gas constant.
• $\text{T}$ is the temperature on absolute scale.

Rearrange the above equation as follows:

  ${\text{n}}_{\text{i}}=\frac{{\text{P}}_{\text{i}}\text{V}}{\text{RT}}$

Mass of i-th component gas present in a gas mixture is calculated as follows:

  ${\text{W}}_{\text{i}}=\left({\text{n}}_{\text{i}}\right)\left({\text{M}}_{\text{i}}\right)$

Where,

• ${\text{W}}_{\text{i}}$ is the mass of i-th component of gas present in a gas mixture.
• ${\text{n}}_{\text{i}}$ is the moles of i-th component gas present.
• ${\text{M}}_{\text{i}}$ is the molar mass of i-th component gas.

Answer to Problem 123AE

Mass of acrylonitrile formed is $1.61\times {10}^3\text{g}$ .

Explanation of Solution

Given Information:The volume of reactor is $150\text{L}$ , temperature of reactor is $25°\text{C}$ , partial pressure of ethane is $0.5\text{MPa}$ , partial pressure of ammonia is $0.8\text{MPa}$ , partial pressure of oxygen is $1.5\text{MPa}$ .

The reaction of the production of acrylonitrile is expressed as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{(g)}+{\text{NH}}_{\text{3}}+\frac{3}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}\text{(g)}+3{\text{H}}_{\text{2}}\text{O}\text{(g)}$

Mole of ethane gas present in a gas mixture, at fixed partial pressure and temperature, is calculated by the equation as follows:

  ${\text{n}}_{\text{ethane}}=\frac{{\text{P}}_{\text{ethane}}\text{V}}{\text{RT}}$

Where,

• ${\text{P}}_{\text{ethane}}$ is the partial pressure of ethane gas.
• $\text{V}$ is the volume of reactor.
• ${\text{n}}_{\text{ethane}}$ is the mole of ethane gas present in container.
• $\text{R}$ is the universal gas constant.
• $\text{T}$ is the temperature on absolute scale.

Conversion of $\text{25 }°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =\text{25 }°\text{C}+273\\ =298\text{K}\end{array}$

Partial pressure of ethane is $0.5\text{MPa}$ .

Volume of reactor is $150\text{L}$ .

Temperature on absolute scale is $298\text{K}$ .

Universal gas constant is $0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$ .

Substitute these values in above equation.

  $\begin{array}{c}{\text{n}}_{\text{ethane}}=\frac{{\text{P}}_{\text{ethane}}\text{V}}{\text{RT}}\\ =\frac{\left(0.5\text{MPa}\right)\left(150.0\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(298\text{K}\right)}\\ =\frac{\left(\left(0.5\times {10}^6\text{Pa}\right)\left(\frac{\text{9}\text{.869}\times {\text{10}}^{-6}\text{atm}}{1\text{Pa}}\right)\right)\left(150.0\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(298\text{K}\right)}\\ =30.29\text{mol}\end{array}$

Moles of acrylonitrileproduced from $30.29\text{mol}$ of ethane gasare calculated as follows:

  $\begin{array}{c}{\text{n}}_{\text{acrylonitrile}}=\left({\text{n}}_{\text{ethane}}\right)\\ =30.29\text{mol}\end{array}$

Mole of ammonia gas present in a gas mixture, at fixed partial pressure and temperature, is calculated by the equation as follows:

  ${\text{n}}_{\text{ammonia}}=\frac{{\text{P}}_{\text{ammonia}}\text{V}}{\text{RT}}$

Where,

• ${\text{P}}_{\text{ammonia}}$ is the partial pressure of ammonia gas.
• $\text{V}$ is the volume of reactor.
• ${\text{n}}_{\text{ammonia}}$ is the mole of ammonia gas present in container.
• $\text{R}$ is the universal gas constant.
• $\text{T}$ is the temperature on absolute scale.

Partial pressure of ammonia is $0.8\text{MPa}$ .

Volume of reactor is $150\text{L}$ .

Temperature on absolute scale is $298\text{K}$ .

Universal gas constant is $0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$ .

Substitute these values in above equation.

  $\begin{array}{c}{\text{n}}_{\text{ammonia}}=\frac{{\text{P}}_{\text{ammonia}}\text{V}}{\text{RT}}\\ =\frac{\left(0.8\text{MPa}\right)\left(150.0\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(298\text{K}\right)}\\ =\frac{\left(0.8\times {10}^6\text{Pa}\left(\frac{\text{9}\text{.869}\times {\text{10}}^{-6}\text{atm}}{1\text{Pa}}\right)\right)\left(150.0\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(298\text{K}\right)}\\ =48.46\text{mol}\end{array}$

Moles of acrylonitrileproducedfrom $48.46\text{mol}$ of ammonia gasare calculated as follows:

  $\begin{array}{c}{\text{n}}_{\text{acrylonitrile}}=\left({\text{n}}_{\text{ammonia}}\right)\\ =48.46\text{mol}\end{array}$

Mole of oxygen gas present in a gas mixture, at fixed partial pressure and temperature is calculated by the equation as follows:

  ${\text{n}}_{\text{oxygen}}=\frac{{\text{P}}_{\text{oxygen}}\text{V}}{\text{RT}}$

Where,

• ${\text{P}}_{\text{oxygen}}$ is the partial pressure of oxygen gas.
• $\text{V}$ is the volume of reactor.
• ${\text{n}}_{\text{oxygen}}$ is the mole of oxygen gas present in container.
• $\text{R}$ is the universal gas constant.
• $\text{T}$ is the temperature on absolute scale.

Partial pressure of oxygen is $1.5\text{MPa}$ .

Volume of reactor is $150\text{L}$ .

Temperature on absolute scale is $298\text{K}$ .

Universal gas constant is $0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$ .

Substitute these values in above equation.

  $\begin{array}{c}{\text{n}}_{\text{oxygen}}=\frac{{\text{P}}_{\text{oxygen}}\text{V}}{\text{RT}}\\ =\frac{\left(1.5\text{MPa}\right)\left(150.0\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(298\text{K}\right)}\\ =\frac{\left(\left(1.5\times {10}^6\text{Pa}\right)\left(\frac{\text{9}\text{.869}\times {\text{10}}^{-6}\text{atm}}{1\text{Pa}}\right)\right)\left(150.0\text{L}\right)}{\left(0.082\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(298\text{K}\right)}\\ =90.87\text{mol}\end{array}$

Moles of acrylonitrileproduced from $90.87\text{mol}$ of oxygen gas are calculated as follows:

  $\begin{array}{c}{\text{n}}_{\text{acrylonitrile}}=\left(\frac{2}{3}\right)\left({\text{n}}_{\text{oxygen}}\right)\\ =\left(\frac{2}{3}\right)\left(90.87\text{mol}\right)\\ =60.58\text{mol}\end{array}$

Moles ofacrylonitrileproduced are minimum in case of ethane so ethane is a limiting reactant.

Mass of acrylonitrile produced is calculated as follows:

  $\begin{array}{c}\text{Mass of acrylonitrile}=\left({\text{n}}_{\text{acrylonitrile}}\right)\left({\text{M}}_{\text{acrylonitrile}}\right)\\ =\left(30.29\text{mol}\right)\left(53.06\text{g}{\text{mol}}^{\text{-1}}\right)\\ =1.61\times {10}^3\text{g}\end{array}$

Hence, mass of acrylonitrile formed is $1.61\times {10}^3\text{g}$ .

Conclusion

Mass of acrylonitrile formed is $1.61\times {10}^3\text{g}$ .