Chapter 3, Problem 3C.9E

(a)

Interpretation Introduction

Interpretation:

Volume of hydrogen needed to produce $1.0{\text{ t NH}}_3$ has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of a gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3C.9E

Volume of hydrogen needed to produce $1.0{\text{ t NH}}_3$ is $3.0\times {10}^5\text{ L}$.

Explanation of Solution

Balanced reaction of Haber process is as follows:

  ${\text{N}}_{\text{2}}+{\text{3H}}_{\text{2}}\to 2{\text{NH}}_3$

Quantity of moles of $1.0{\text{ t NH}}_3$ can be calculated as follows:

  $\begin{array}{c}\text{Mole}\left({\text{NH}}_3\right)=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\left(\frac{1.0\text{ t}}{\text{17}\text{.03 g/mol}}\right)\left(\frac{{10}^3\text{ kg}}{1\text{ t}}\right)\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)\\ =5.9\times {10}^4\text{ mol}\end{array}$

According to reaction, 3 moles of ${\text{H}}_2$ gas produces 2 moles of ${\text{NH}}_3$ thus amount of ${\text{H}}_2$ required to produce $5.9\times {10}^4\text{ mol}$ of ${\text{NH}}_3$ can be calculated as follows:

  $\begin{array}{c}\text{Mass}\left({\text{H}}_{\text{2}}\right)=\text{Mole}\left({\text{NH}}_{\text{3}}\right)\left(\frac{3{\text{ mol H}}_2}{2{\text{ mol NH}}_3}\right)\\ =\left(5.9\times {10}^4{\text{ mol NH}}_{\text{3}}\right)\left(\frac{3{\text{ mol H}}_2}{2{\text{ mol NH}}_3}\right)\\ =8.8\times {10}^4{\text{ mol H}}_2\end{array}$

The expression to calculate volume of gas is as follows:

  $V=\frac{nRT}{P}$        (1)

Conversion factor to convert $350\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =350\text{°C}+273.15\\ =623.15\text{K}\end{array}$

Substitute $15.00\text{ atm}$ for $P$, $8.8\times {10}^4\text{ mol}$ for $n$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $623.15\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}V=\frac{\left(8.8\times {10}^4\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(623.15\text{K}\right)}{\left(15.00\text{ atm}\right)}\\ =3.0\times {10}^5\text{ L}\end{array}$

Hence, volume of hydrogen needed to produce $1.0{\text{ t NH}}_3$ is $3.0\times {10}^5\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Volume of hydrogen needed to produce $1.0{\text{ t NH}}_3$ at $376\text{ atm}$ and $250°\text{C}$ has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of a gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3C.9E

Volume of hydrogen needed to produce $1.0{\text{ t NH}}_3$ at $376\text{ atm}$ and $250°\text{C}$ is $1.0\times {10}^4\text{ L}$.

Explanation of Solution

The expression to calculate volume of gas is as follows:

  $V=\frac{nRT}{P}$        (1)

Conversion factor to convert $250\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =250\text{°C}+273.15\\ =523.15\text{K}\end{array}$

Substitute $376\text{ atm}$ for $P$, $8.8\times {10}^4\text{ mol}$ for $n$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $523.15\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}V=\frac{\left(8.8\times {10}^4\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(523.15\text{K}\right)}{\left(376\text{ atm}\right)}\\ =1.0\times {10}^4\text{ L}\end{array}$

Hence volume of hydrogen needed to produce $1.0{\text{ t NH}}_3$ at $376\text{ atm}$ and $250°\text{C}$ is $1.0\times {10}^4\text{ L}$.