Chapter 14, Problem 42E

Interpretation Introduction

Interpretation:

In the reaction, $4{\text{NH}}_{\text{3}}\left(\text{g}\right)+5{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$, when $\text{6}\text{.98}\text{L}$ of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ gas at $\text{129}{°}^{}\text{C}$ and $\text{0}\text{.836}\text{atm}$ is reacted, the volume of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ produced when it is collected at $\text{74}{°}^{}\text{C}$ and $\text{0}\text{.630}\text{atm}$ is to be calculated.

Concept introduction:

Ideal gas is defined as the gas in which the collisions between the molecules and the atoms are perfectly elastic and there are no intermolecular attractive forces found between them. The ideal gas equation is given by the expression as shown below

$\text{PV}=\text{nRT}$

Answer to Problem 42E

When $\text{6}\text{.98}\text{L}$ of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ gas at $\text{129}{°}^{}\text{C}$ and $\text{0}\text{.836}\text{atm}$ is reacted, the volume of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ produced if it is collected at $\text{74}{°}^{}\text{C}$ and $\text{0}\text{.630}\text{atm}$ is $\text{9}\text{.6}\text{L}$.

Explanation of Solution

In the reaction, $4{\text{NH}}_{\text{3}}\left(\text{g}\right)+5{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$, the initial pressure and temperature of ${\text{O}}_{\text{2}}$ are $\text{0}\text{.836}\text{atm}$ and $\text{129}{°}^{}\text{C}$ respectively and the initial volume of ${\text{O}}_{\text{2}}$ is $\text{6}\text{.98}\text{L}$. The final temperature and pressure of ${\text{O}}_{\text{2}}$ are $\text{74}{°}^{}\text{C}$ and $\text{0}\text{.630}\text{atm}$ respectively.

Conversion of temperature from Celsius to Kelvin can be done as shown below.

$\text{T}\left(\text{K}\right)=\text{T}\left({}^{\text{o}}\text{C}\right)+273$

The final temperature is converted into Kelvin as shown below.

$\begin{array}{rll}{\text{T}}_{\text{final}}\left(\text{K}\right)& =& \left(\text{74}+273\right)\text{K}\\ & =& \text{347}\text{K}\end{array}$

The initial temperature is converted into Kelvin as shown below.

$\begin{array}{rll}{\text{T}}_{\text{initial}}\left(\text{K}\right)& =& \left(\text{129}+273\right)\text{K}\\ & =& \text{402}\text{K}\end{array}$

Therefore, the initial and final temperature of ${\text{O}}_{\text{2}}$ are $\text{402}\text{K}$ and $\text{347}\text{K}$ respectively.

The relation between the initial and final pressure, volume and temperature of gas is shown below.

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}=\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$ …(1)

Where,

• ${\text{P}}_{\text{1}}$ is the initial pressure.

• ${\text{V}}_{\text{1}}$ is the initial volume.

• ${\text{T}}_{\text{1}}$ is the initial temperature.

• ${\text{P}}_{\text{2}}$ is the final pressure.

• ${\text{V}}_{\text{2}}$ is the final volume.

• ${\text{T}}_{\text{2}}$ is the final temperature.

Substitute the values of final and initial temperature, pressure and volume into the equation (1).

$\begin{array}{rll}\frac{\text{0}\text{.836}\text{atm}\times \text{6}\text{.98}\text{L}}{402\text{K}}& =& \frac{\text{0}\text{.630}\text{atm}\times {\text{V}}_{\text{2}}}{\text{347}\text{K}}\\ {\text{V}}_{\text{2}}& =& \frac{\text{0}\text{.836}\text{atm}\times \text{6}\text{.98}\text{L}\times \text{347}\text{K}}{402\text{K}\times \text{0}\text{.630}\text{atm}}\\ & =& 8\text{L}\end{array}$

Therefore, the final volume of ${\text{O}}_{\text{2}}$ gas is $8\text{L}$.

The volume of ${\text{O}}_{\text{2}}$ that produces $\text{6}\text{L}{\text{H}}_{\text{2}}\text{O}$ is $\text{5}\text{L}$.

Therefore the volume of ${\text{H}}_{\text{2}}\text{O}\left({\text{V}}_{{\text{H}}_{\text{2}}\text{O}}\right)$ that are produced by $8.00\text{L}$ of ${\text{O}}_{\text{2}}$ is calculated as shown below.

$\begin{array}{rll}{\text{V}}_{{\text{H}}_{\text{2}}\text{O}}& =& \frac{\text{6}\text{L}{\text{H}}_2\text{O}}{\text{5}\text{L}{\text{O}}_{\text{2}}}\times 8.00\text{L}{\text{O}}_{\text{2}}\\ & =& 9.6\text{L}{\text{H}}_2\text{O}\end{array}$

Therefore, the volume of ${\text{H}}_{\text{2}}\text{O}$ gas produced is $9.6\text{L}$.

Conclusion

When $\text{6}\text{.98}\text{L}$ of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ gas at $\text{129}{°}^{}\text{C}$ and $\text{0}\text{.836}\text{atm}$ is reacted, the volume of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ produced if it is collected at $\text{74}{°}^{}\text{C}$ and $\text{0}\text{.630}\text{atm}$ is $\text{9}\text{.6}\text{L}$.