Chapter 6, Problem 6.115QP

Interpretation Introduction

Interpretation: The percent decrease in pressure of a sealed reaction vessel during the reaction between $\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{H}}_{\text{2}}$ and $\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{N}}_{\text{2}}$ if half of the ${\text{N}}_{\text{2}}$ consumed is to be calculated.

Concept introduction: The reaction between nitrogen and hydrogen involves the formation of ammonia. When one mole of nitrogen reacts with three moles of hydrogen then it produces 2 moles of ammonia. The balanced equation is given below,

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

From the ideal gas equation, at constant temperature and volume, Pressure is directly proportional to the number of moles,

$P\propto n$

To determine: The percent decrease in pressure of a sealed reaction vessel during the reaction between $\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{H}}_{\text{2}}$ and $\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{N}}_{\text{2}}$ if half of ${\text{N}}_{\text{2}}$ consumes.

Answer to Problem 6.115QP

Solution

The percent decrease in pressure of a sealed reaction vessel during the reaction between $\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{H}}_{\text{2}}$ and $\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{N}}_{\text{2}}$ if half of the ${\text{N}}_{\text{2}}$ consumes is $25\%$.

Explanation of Solution

Explanation

Given

The number of moles of ${\text{H}}_{\text{2}}$ is $\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}$.

The number of moles of ${\text{N}}_{\text{2}}$ is $\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}$.

So, the total number of moles that are present initially is,

$\begin{array}{c}{n}_{\text{i}}=\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}+\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}\\ =\text{4}\text{.80}\times {\text{10}}^{\text{3}}\text{mol}\end{array}$

The balanced chemical equation between ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ to form ${\text{NH}}_{\text{3}}$ is shown below,

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

According to this equation, $\text{3}\text{moles}$ of ${\text{H}}_{\text{2}}$ are consumed by $1\text{mole}$ of ${\text{N}}_2$ to make $\text{2}\text{moles}$ of ${\text{NH}}_{\text{3}}$.

$\begin{array}{c}1\text{mole}\text{of}{\text{N}}_{\text{2}}=3\text{mol}\text{of}{\text{H}}_{\text{2}}\\ \text{1}\text{.20}\times {\text{10}}^3\text{mole}\text{of}{\text{N}}_{\text{2}}=\text{3}\times \text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}\text{of}{\text{H}}_{\text{2}}\\ =3.60\times {10}^3\text{mol}\text{of}{\text{H}}_{\text{2}}\end{array}$

During the reaction, half of the moles of ${\text{N}}_{\text{2}}$ are consumed.

$\begin{array}{c}1.20\times {10}^3\text{mole}\text{of}{\text{N}}_{\text{2}}=\frac{3.60\times {\text{10}}^{\text{3}}\text{mol}{\text{H}}_{\text{2}}}{2}\\ =\text{1}\text{.80}\times {\text{10}}^{\text{3}}\text{mol}{\text{H}}_{\text{2}}\end{array}$

$\begin{array}{c}\text{1}\text{mole}\text{of}{\text{N}}_2=2\text{mole}\text{of}{\text{NH}}_{\text{3}}\\ \text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mole}\text{of}{\text{N}}_{\text{2}}=\text{2}\times \text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mole}\text{of}{\text{NH}}_{\text{3}}\\ =2.40\times {10}^3\text{mole}\text{of}{\text{NH}}_{\text{3}}\end{array}$

Therefore, the final number of moles $\left(n_{\text{f}}\right)$ is,

$\begin{array}{c}{n}_{\text{f}}=\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}{\text{H}}_{\text{2}}+\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}{\text{N}}_{\text{2}}+\text{2}\text{.40}\times {\text{10}}^{\text{3}}\text{mol}{\text{NH}}_{\text{3}}\\ =\text{7}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}\end{array}$

Since, half of the ${\text{N}}_{\text{2}}$ is consumed. Therefore,

$\begin{array}{c}{n}_f=\frac{7.20\times {10}^3\text{mol}}{2}\\ =\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}\end{array}$

Since, $P\propto n$.

The percentage decrease in pressure is calculated by the formula,

% decrease in pressure $\begin{array}{l}=\frac{P_f-P_i}{P_i}\times 100\%\\ =\frac{n_f-n_i}{n_i}\times 100\%\end{array}$

Substitute the values of $n_f$ and $n_i$ in the above equation,

$\begin{array}{c}\%\text{decrease}\text{in}\text{pressure}=\frac{P_f-P_i}{P_i}\times 100\%\\ =\frac{n_f-n_i}{n_i}\times 100\%\\ =\frac{4.80\times {10}^3\text{mol}-3.60\times {10}^3\text{mol}}{4.80\times {10}^3\text{mol}}\\ =25\%\end{array}$

Conclusion

The percent decrease in pressure of a sealed reaction vessel during the reaction between $\text{3}\text{.60}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{H}}_{\text{2}}$ and $\text{1}\text{.20}\times {\text{10}}^{\text{3}}\text{mol}$ of ${\text{N}}_{\text{2}}$ if half of the ${\text{N}}_{\text{2}}$ consumes is $25\%$