Chapter 5, Problem 5.148P

(a)

Interpretation Introduction

Interpretation:

The volumes of ${\text{NH}}_{\text{3}}$ needed per liter of flue gas at $1\text{ atm}$ are to be calculated

Concept introduction:

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

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

Answer to Problem 5.148P

The volumes of ${\text{NH}}_{\text{3}}$ are needed per liter of flue gas at $1\text{ atm}$ are $4.5\times {10}^{-5}\text{ L}$.

Explanation of Solution

The equation for the reaction of ${\text{NH}}_{\text{3}}$, $\text{NO}$ with ${\text{O}}_{\text{2}}$ is as follows:

${\text{4NH}}_3\left(g\right)\text{ + 4NO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{N}}_2\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$ (1)

The formula to convert $°\text{C}$ to Kelvin is:

$T_{\text{K}}=T_{\text{°C}}+273.15$ (2)

Substitute $\text{365 }°\text{C}$ for $T_{\text{°C}}$ in equation (2).

$\begin{array}{c}{T}_{\text{K}}=365+273.15\\ =638.15\text{ K}\end{array}$

The expression to calculate the moles of the $\text{NO}$ is as follows,

$PV=nRT$ (3)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the $\text{NO}$ and $R$ is the gas constant.

Rearrange the equation (3) to calculate $n$ as follows,

$n=\frac{PV}{RT}$                                                                                                                       (4)

Substitute the value $\text{4}\text{.5}\times {\text{10}}^{-5}\text{ atm}$ for $P$, $\text{638}\text{.15 K}$ for $T$, $\text{1}\text{.0 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (4).

$\begin{array}{c}n=\frac{\left(1.0\text{ L}\right)\left(\text{4}\text{.5}\times {\text{10}}^{-5}\text{ atm}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(638.15\text{ K}\right)}\\ =8.589\times {10}^{-7}\text{ mol}\end{array}$

From the equation (1), four moles of ${\text{NH}}_{\text{3}}$ reacts with four moles of $\text{NO}$ so, the moles of ${\text{NH}}_{\text{3}}$ is calculated from $\text{NO}$ is as follows:

${\text{Moles of NH}}_{\text{3}}=\left(\text{Moles of NO}\right)\left(\frac{{\text{2 mol of NH}}_{\text{3}}}{\text{2 mol of NO}}\right)$ (5)

Substitute the value $8.589\times {10}^{-7}\text{ mol}$ for $\text{NO}$ in the equation (5)

$\begin{array}{c}{\text{Moles of NH}}_{\text{3}}=\left(8.589\times {10}^{-7}\text{ mol}\right)\left(\frac{{\text{2 mol of NH}}_{\text{3}}}{\text{2 mol of NO}}\right)\\ =8.589\times {10}^{-7}\text{ mol}\end{array}$

The expression to calculate the volume is as follows,

$PV=nRT$ (3)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the $\text{NO}$ and $R$ is the gas constant.

Rearrange the equation (3) to calculate $V$ as follows,

$V=\frac{nRT}{P}$                                                                                                                     (6)

Substitute the value $\text{1 atm}$ for $P$, $\text{638}\text{.15 K}$ for $T$, $8.5891\times {10}^{-7}\text{ mol}$ for $n$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (6).

$\begin{array}{c}V=\frac{\left(\text{8}\text{.5891}\times {\text{10}}^{-7}\text{ atm}\right)\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(638.15\text{ K}\right)}{\left(1.0\text{ atm}\right)}\\ =4.50001\times {10}^{-5}\text{ L}\\ \approx 4.5\times {10}^{-5}\text{ L}\end{array}$

Conclusion

The volumes of ${\text{NH}}_{\text{3}}$ are needed per liter of flue gas at $1\text{ atm}$ are $4.5\times {10}^{-5}\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

The mass of ${\text{NH}}_{\text{3}}$ are needed per kiloliter at $365°\text{C}$ and $1\text{ atm}$ is to be calculated.

Concept introduction:

The relationship between pressure and volume can be expressed as follows,

$PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

$V\alpha T$

Here, $T$ is the temperature and $V$ is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

$\text{V }\alpha n$

Here, $n$ is the mole of the gas and $V$ is the volume.

The ideal gas equation can be expressed as follows,

$PV=nRT$

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

Answer to Problem 5.148P

The mass of ${\text{NH}}_{\text{3}}$ are needed per kiloliter at $365°\text{C}$ and $1\text{ atm}$ is $0.015\text{ g}$.

Explanation of Solution

Substitute the value $\text{4}\text{.5}\times {\text{10}}^{-5}\text{ atm}$ for $P$, $\text{638}\text{.15 K}$ for $T$, $\text{1}\text{.0 kL}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (4).

$\begin{array}{c}n=\frac{\left(1.0\text{ kL}\left(\frac{1000\text{ L}}{1\text{ kL}}\right)\right)\left(\text{4}\text{.5}\times {\text{10}}^{-5}\text{ atm}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(638.15\text{ K}\right)}\\ =8.589\times {10}^{-4}\text{ mol}\end{array}$

From the equation (1), four moles of ${\text{NH}}_{\text{3}}$ reacts with four moles of $\text{NO}$ so, the moles of ${\text{NH}}_{\text{3}}$ is calculated from $\text{NO}$ is as follows:

${\text{Moles of NH}}_{\text{3}}=\left(\text{Moles of NO}\right)\left(\frac{{\text{2 mol of NH}}_{\text{3}}}{\text{2 mol of NO}}\right)$ (7)

Substitute the value $8.589\times {10}^{-4}\text{ mol}$ for $\text{NO}$ in the equation (7)

$\begin{array}{c}{\text{Moles of NH}}_{\text{3}}=\left(8.589\times {10}^{-4}\text{ mol}\right)\left(\frac{{\text{2 mol of NH}}_{\text{3}}}{\text{2 mol of NO}}\right)\\ =8.589\times {10}^{-4}\text{ mol}\end{array}$

The expression to calculate the mass of ${\text{NH}}_{\text{3}}$ is as follows:

${\text{Moles of NH}}_{\text{3}}=\frac{{\text{Mass of NH}}_{\text{3}}}{{\text{Molar mass of NH}}_{\text{3}}}$ (8)

Rearrange the equation (8) to calculate the mass of ${\text{NH}}_{\text{3}}$ as follows,

${\text{Mass of NH}}_{\text{3}}=\left({\text{Moles of NH}}_{\text{3}}\right)\left({\text{Molar mass of NH}}_{\text{3}}\right)$                                              (9)

Substitute the value $8.589\times {10}^{-4}\text{ mol}$ for moles of ${\text{NH}}_{\text{3}}$ and $17.03\text{ g/mol}$ for the molar mass of ${\text{NH}}_{\text{3}}$ in the equation (9).

$\begin{array}{c}{\text{Mass of NH}}_{\text{3}}=\left(8.589\times {10}^{-4}\text{ mol}\right)\left(17.03\text{ g/mol}\right)\\ =0.014627\text{ g}\\ =\text{0}\text{.015 g}\end{array}$

Conclusion

The mass of ${\text{NH}}_{\text{3}}$ are needed per kiloliter at $365°\text{C}$ and $1\text{ atm}$ is $0.015\text{ g}$.