Chapter 5, Problem 99E

Interpretation Introduction

Interpretation: The total pressure of given reaction tank and the partial pressures of nitrogen gas at ${\text{127}}^{\text{o}}\text{C}$ in the given reaction tank of $\text{250}\text{L}$ are needed to be determined if the mass of dimethyl hydrazine is $\text{150}\text{g}$.

Concept introduction:

• Number of moles of a substance,

  From its given mass is,

  $\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

  Pressure of the gas, according to ideal gas equation,

  $\text{Pressure}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\text{×}\text{Temperature}}{\text{Volume}}$

• Total pressure of a tank containing mixture of gases is the sum of individual partial pressures of constituted gases.

Answer to Problem 99E

Answer

The partial pressure of ${\text{N}}_{\text{2}}$ in the given reaction is $0.98\text{atm}$.

Total pressure of given reaction container is $2.93\text{atm}$.

Explanation of Solution

Explanation

• To find: the number of moles of dimethyl hydrazine from its given mass.

The number of moles of dimethyl hydrazine is $\text{2}\text{.5}\text{mol}$.

The given mass for dimethyl hydrazine is $\text{150}\text{g}$.

Number of moles of a substance from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore, the number of moles of dimethyl hydrazine in the flask is,

$\text{Number of moles of }dimethylhydrazine\text{=}\frac{\text{150}\text{g}}{60.1\text{g/mol}}\text{=}2.5\text{mol}$

• To find: the molar ratio between produced ${\text{N}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ with the reactant dimethyl hydrazine and the number of moles of each produced gases in the given reaction

3 moles of ${\text{N}}_{\text{2}}$ is produced by 1 mole of dimethyl hydrazine in the given reaction.

4 moles of ${\text{H}}_{\text{2}}\text{O}$ is produced by 1 mole of dimethyl hydrazine in the given reaction.

2 moles of ${\text{CO}}_{\text{2}}$ is produced by 1 mole of dimethyl hydrazine in the given reaction.

The number of moles of produced ${\text{N}}_{\text{2}}$ is $\text{7}\text{.5}\text{mol}$.

The number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ is $10\text{mol}$.

The number of moles of produced ${\text{CO}}_{\text{2}}$ is $\text{5}\text{mol}$.

For ${\text{N}}_{\text{2}}$,

• The balanced equation of the reaction is given as ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}\text{+}\text{2}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\to \text{3}{\text{N}}_{\text{2}}\text{+}\text{4}{\text{H}}_{\text{2}}\text{O}\text{+}\text{2}{\text{CO}}_{\text{2}}$.

  Here the 1 moles of dimethyl hydrazine is producing 3 moles of ${\text{N}}_{\text{2}}$.

  That is,

  The mole ratio between dimethyl hydrazine and ${\text{N}}_{\text{2}}$ is,

$1:3$

• The number of moles of reacted dimethyl hydrazine is $\text{2}\text{.5}\text{mol}$.

The mole ratio between dimethyl hydrazine and produced ${\text{N}}_{\text{2}}$ is 1:3.

Therefore,

The number of moles of produced ${\text{N}}_{\text{2}}$ is,

$\text{2}\text{.5}\text{mol}\text{×}\frac{\text{3}\text{mol}{\text{N}}_{\text{2}}}{\text{1}\text{mol}\text{dimethyl}\text{hydrazine}}\text{=}\text{7}\text{.5}\text{mol}$

Hence, the number of moles of produced ${\text{N}}_{\text{2}}$ is $\text{7}\text{.5}\text{mol}$.

For ${\text{H}}_{\text{2}}\text{O}$,

• The balanced equation of the reaction is given as ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}\text{+}\text{2}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\to \text{3}{\text{N}}_{\text{2}}\text{+}\text{4}{\text{H}}_{\text{2}}\text{O}\text{+}\text{2}{\text{CO}}_{\text{2}}$.

  Here the 1 moles of dimethyl hydrazine is producing 4 moles of ${\text{H}}_{\text{2}}\text{O}$.

  That is,

  The mole ratio between dimethyl hydrazine and ${\text{H}}_{\text{2}}\text{O}$ is,

$1:4$

• The number of moles of reacted dimethyl hydrazine is $\text{2}\text{.5}\text{mol}$.

The mole ratio between dimethyl hydrazine and produced ${\text{H}}_{\text{2}}\text{O}$ is 1:3.

Therefore,

The number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ is,

$\text{2}\text{.5}\text{mol}\text{×}\frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}\text{dimethyl}\text{hydrazine}}\text{=}10\text{mol}$

Hence, the number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ is $10\text{mol}$.

For ${\text{CO}}_{\text{2}}$,

• The balanced equation of the reaction is given as ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}\text{+}\text{2}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\to \text{3}{\text{N}}_{\text{2}}\text{+}\text{4}{\text{H}}_{\text{2}}\text{O}\text{+}\text{2}{\text{CO}}_{\text{2}}$.

  Here the 1 moles of dimethyl hydrazine is producing 2 moles of ${\text{CO}}_{\text{2}}$.

  That is,

  The mole ratio between dimethyl hydrazine and ${\text{CO}}_{\text{2}}$ is,

$1:2$

• The number of moles of reacted dimethyl hydrazine is $\text{2}\text{.5}\text{mol}$.

The mole ratio between dimethyl hydrazine and produced ${\text{CO}}_{\text{2}}$ is 1:2.

Therefore,

The number of moles of produced ${\text{CO}}_{\text{2}}$ is,

$\text{2}\text{.5}\text{mol}\text{×}\frac{2\text{mol}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}\text{dimethyl}\text{hydrazine}}\text{=}\text{5}\text{mol}$

Hence, the number of moles of produced ${\text{N}}_{\text{2}}$ is $\text{5}\text{mol}$.

• To find: the partial pressures nitrogen gas in the given reaction.

The partial pressure of ${\text{N}}_{\text{2}}$ in the given reaction is $0.98\text{atm}$.

The number of moles of produced ${\text{N}}_{\text{2}}$ is calculated as $\text{7}\text{.5}\text{mol}$.

The temperature of reaction is given as ${\text{127}}^{\text{o}}\text{C}\text{=}\text{(127}\text{+}\text{273}\text{)K}\text{=}\text{400}\text{K}$

The volume of tank is given as $\text{250}\text{L}$.

Pressure of a gas, according to ideal gas equation,

$\text{Pressure}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\text{×}\text{Temperature}}{\text{Volume}}$

Therefore,

The partial pressure of ${\text{N}}_{\text{2}}$ in the given reaction is,

$\begin{array}{l}{\text{P}}_{N_2}\text{=}\frac{\text{7}\text{.5}\text{mol}\text{×}\text{0}\text{.08206}\text{L}\text{atm}\text{/}\text{K}\text{mol}\text{×}\text{400}\text{K}}{\text{250}\text{L}}\\ \text{=}\text{0}\text{.98}\text{atm}\end{array}$

Hence,

The partial pressure of ${\text{N}}_{\text{2}}$ in the given reaction is $0.98\text{atm}$.

• To find: the partial pressures of ${\text{H}}_{\text{2}}\text{O}$ in the given reaction.

The partial pressure of ${\text{H}}_{\text{2}}\text{O}$ in the given reaction is $1.3\text{atm}$.

The number of moles of produced ${\text{H}}_{\text{2}}\text{O}$ is calculated as $\text{10}\text{mol}$.

The temperature of reaction is given as ${\text{127}}^{\text{o}}\text{C}\text{=}\text{(127}\text{+}\text{273}\text{)K}\text{=}\text{400}\text{K}$

The volume of tank is given as $\text{250}\text{L}$.

Pressure of a gas, according to ideal gas equation,

$\text{Pressure}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\text{×}\text{Temperature}}{\text{Volume}}$

Therefore,

The partial pressure of ${\text{H}}_{\text{2}}\text{O}$ in the given reaction is,

$\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\frac{\text{10}\text{mol}\text{×}\text{0}\text{.08206}\text{L}\text{atm}\text{/}\text{K}\text{mol}\text{×}\text{400}\text{K}}{\text{250}\text{L}}\\ \text{=}1.3\text{atm}\end{array}$

Hence,

The partial pressure of ${\text{H}}_{\text{2}}\text{O}$ in the given reaction is $1.3\text{atm}$.

• To find: the partial pressures of ${\text{CO}}_{\text{2}}$ in the given reaction.

The partial pressure of ${\text{CO}}_{\text{2}}$ in the given reaction is $0.65\text{atm}$.

The number of moles of produced ${\text{CO}}_{\text{2}}$ is calculated as $\text{5}\text{mol}$.

The temperature of reaction is given as ${\text{127}}^{\text{o}}\text{C}\text{=}\text{(127}\text{+}\text{273}\text{)K}\text{=}\text{400}\text{K}$

The volume of tank is given as $\text{250}\text{L}$.

Pressure of a gas, according to ideal gas equation,

$\text{Pressure}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\text{×}\text{Temperature}}{\text{Volume}}$

Therefore,

The partial pressure of ${\text{CO}}_{\text{2}}$ in the given reaction is,

$\begin{array}{l}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{5}\text{mol}\text{×}\text{0}\text{.08206}\text{L}\text{atm}\text{/}\text{K}\text{mol}\text{×}\text{400}\text{K}}{\text{250}\text{L}}\\ \text{=}\text{0}\text{.65}\text{atm}\end{array}$

Hence,

The partial pressure of ${\text{CO}}_{\text{2}}$ in the given reaction is $0.65\text{atm}$.

• To find: the total pressure of the given reaction tank.

The total pressure of the given reaction tank is $2.93\text{atm}$.

The total pressure of a tank containing mixture of gases is the sum of individual partial pressures of constituted gases.

The partial pressures of products in given reaction are calculated above.

Therefore,

The total pressure of the given reaction tank is

${\text{P}}_{{\text{N}}_{\text{2}}}\text{+}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{+}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{}\text{=}\text{0}\text{.98}\text{atm}\text{+}\text{1}\text{.3}\text{atm}\text{+}\text{0}\text{.65}\text{atm}\text{=}\text{2}\text{.93}\text{atm}$

Hence, the final total pressure of the given reaction tank is $2.93\text{atm}$.

Conclusion

Conclusion

The partial pressures of ${\text{N}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ gases in the given reaction is determined by using ideal gas equation. The total pressure of container is determined by adding the partial pressures of ${\text{N}}_{\text{2}}$ ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ gases.