Chapter 10, Problem 98AP

Interpretation Introduction

Interpretation:

The total volume and the partial pressures of the gases formed from the decomposition of nitroglycerin are to be calculated.

Concept Introduction:

The expression foran ideal gas equation is

$PV=nRT.$

Here, $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is a gas constant, and $T$ is temperature.

The expression to calculate mole fraction $\left({\chi }_i\right)$ is

${\chi }_i=\frac{n_i}{n_{\text{total}}}.$

Here, $n_i$ is the moles of component and $n_{\text{total}}$ is the total number of moles.

The relation between partial pressure $\left(P_i\right)$ and total pressure $\left(P_{\text{total}}\right)$ is:

${\chi }_i=\frac{P_i}{P_{\text{total}}}.$

The number of moles is given by the expression, which is as:

$n=\frac{\text{given mass}\left(m\right)}{\text{molar mass}\left(M\right)}.$

The conversion factor is $\left(\frac{29\text{ mol product}}{4\text{ mol nitroglycerine}}\right)$.

So the moles can be calculated by the expression as:

$n=\frac{\text{given mass}\left(m\right)}{\text{molar mass}\left(M\right)}\times \left(\frac{29\text{ mol product}}{4\text{ mol nitroglycerine}}\right).$

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T\left(\text{in K}\right)=T\left(\text{in }°\text{C}\right)+\mathrm{273.15.}$

Answer to Problem 98AP

Solution: $V=1.7\times {10}^2\text{ L}\text{.}$

$P_{{\text{CO}}_2}=0.49\text{ atm}\text{.}$

$P_{{\text{H}}_2\text{O}}=0.41\text{ atm}\text{.}$

$P_{{\text{N}}_2}=0.25\text{ atm}\text{.}$

$P_{{\text{O}}_2}=0.041\text{ atm}\text{.}$

Explanation of Solution

Given information: $4{\text{C}}_3{\text{H}}_5{\left({\text{NO}}_3\right)}_3\left(s\right)\to 12{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)+6{\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right).$

Pressure is $1.2\text{ atm,}$ temperature is ${25}^o\text{C}\text{.}$

Mass of nitroglycerine is $2.6\times {10}^2\text{ g}\text{.}$

The mass of $1\text{ mole}$ of nitroglycerine is equal to its molar mass, that is $\text{227}\text{.09 g}\text{.}$

$\text{1 mol nitroglycerine}=\text{227}\text{.09 g}\left[\text{molar mass of nitroglycerine}=\text{227}\text{.09 g}/\text{mol }\right].$

Invert the conversion factor, whichbecomes $\left(\frac{\text{1 mol nitroglycerine}}{\text{227}\text{.09 g nitroglycerine}}\right)$.

According to the given reaction, the mole ratio of nitroglycerine and the products is

$4\text{ mol nitroglycerine}=29\text{ mol product}\text{.}$

Therefore, the conversion factor becomes $\left(\frac{29\text{ mol product}}{4\text{ mol nitroglycerine}}\right)$.

Multiply the conversion factors with the mass of nitroglycerine in order to calculate the moles of the products as

The moles can be calculated by the expression as

$n=\frac{\text{given mass}\left(m\right)}{\text{molar mass}\left(M\right)}\times \left(\frac{29\text{ mol product}}{4\text{ mol nitroglycerine}}\right)$

$\begin{array}{c}\text{Mol of products}=\left(\frac{\text{1 }\cancel{\text{mol nitroglycerine}}}{\text{227}\text{.09 }\cancel{\text{g nitroglycerine}}}\right)\times \left(\frac{29\text{ mol product}}{4\cancel{\text{mol nitroglycerine}}}\right)\times 2.6\times {10}^2\cancel{\text{g nitroglycerine}}\\ =8.3\text{ mol}\text{.}\end{array}$

Temperature is in Celsius. Convert it into Kelvin as:

$\begin{array}{c}T=\left(25+273\right)\text{ K}\\ =\text{298 K}\text{.}\end{array}$

The expression foran ideal gas equation is

$PV=nRT.$

Rearrange the above expression for $V:$

$V=\frac{nRT}{P}.$

Substitute $8.3\text{ mol}$ for $n$, $0.0821\text{ L atm/mol-K}$ for $R$, $298\text{ K}$ for $T,$ and $1.2\text{ atm}$ for $P$ in the above expression as

$\begin{array}{c}V=\frac{8.3\cancel{\text{mol}}\times 0.0821\text{ L }\cancel{\text{atm}}\text{/mol-}\cancel{\text{K}}\times 298\cancel{\text{K}}}{1.2\cancel{\text{atm}}}\\ =169.2\text{ L}\text{.}\end{array}$

As the minimum number of significant figures is 2, the final answer should also contain two significant figures, that is, $1.7\times {10}^2\text{ L}$.

The relation between partial pressure and total pressure is

$P_i={\chi }_i{P}_T.$ …… (1)

The formula to calculate the mole fraction is

${\chi }_i=\frac{n_i}{n_{\text{total}}}.$ …… (2)

Now, calculate the mole fractions of each product.

For ${\text{CO}}_2$, substitute $12\text{ mol}$ for $n_i$ and $\text{29 mol}$ for $n_{\text{total}}$ in the above expression as

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

$\begin{array}{c}{\chi }_{{\text{CO}}_2}=\frac{12\cancel{\text{mol}}}{\text{29 }\cancel{\text{mol}}}\\ =\mathrm{0.41.}\end{array}$

For ${\text{H}}_2\text{O}$, substitute $10\text{ mol}$ for $n_i$ and $\text{29 mol}$ for $n_{\text{total}}$ in equation (2) as

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

$\begin{array}{c}{\chi }_{{\text{H}}_2\text{O}}=\frac{10\cancel{\text{mol}}}{\text{29 }\cancel{\text{mol}}}\\ =\mathrm{0.34.}\end{array}$

For ${\text{N}}_2$, substitute $\text{6 mol}$ for $n_i$ and $\text{29 mol}$ for $n_{\text{total}}$ in equation (2) as

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

$\begin{array}{c}{\chi }_{{\text{N}}_2}=\frac{\text{6 }\cancel{\text{mol}}}{\text{29 }\cancel{\text{mol}}}\\ =\mathrm{0.21.}\end{array}$

For ${\text{O}}_2$, substitute $\text{1 mol}$ for $n_i$ and $\text{29 mol}$ for $n_{\text{total}}$ in equation (2) as

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

$\begin{array}{c}{\chi }_{{\text{O}}_2}=\frac{\text{1 }\cancel{\text{mol}}}{\text{29 }\cancel{\text{mol}}}\\ =\mathrm{0.034.}\end{array}$

For ${\text{CO}}_2$, substitute $0.41$ for ${\chi }_i$ and $1.2\text{ atm}$ for $P_T$ in equation (1) as

$P_i={\chi }_i{P}_T$

$\begin{array}{c}{P}_{{\text{CO}}_2}=\left(0.41\right)\left(1.2\text{ atm}\right)\\ =0.49\text{ atm}\text{.}\end{array}$

For ${\text{H}}_2\text{O}$, substitute $0.34$ for ${\chi }_i$ and $1.2\text{ atm}$ for $P_T$ in equation (1) as

$P_i={\chi }_i{P}_T$

$\begin{array}{c}{P}_{{\text{H}}_2\text{O}}=\left(0.34\right)\left(1.2\text{ atm}\right)\\ =0.41\text{ atm}\text{.}\end{array}$

For ${\text{N}}_2$, substitute $0.21$ for ${\chi }_i$ and $1.2\text{ atm}$ for $P_T$ in equation (1) as

$P_i={\chi }_i{P}_T$

$\begin{array}{c}{P}_{{\text{N}}_2}=\left(0.21\right)\left(1.2\text{ atm}\right)\\ =0.25\text{ atm}\text{.}\end{array}$

For ${\text{O}}_2$, substitute $0.034$ for ${\chi }_i$ and $1.2\text{ atm}$ for $P_T$ in equation (1) as

$P_i={\chi }_i{P}_T$

$\begin{array}{c}{P}_{{\text{O}}_2}=\left(0.034\right)\left(1.2\text{ atm}\right)\\ =0.041\text{ atm}\text{.}\end{array}$

Conclusion

The total volume of the gas is $1.7\times {10}^2\text{ L}$. The partial pressures of ${\text{CO}}_2$, ${\text{H}}_2\text{O}$, ${\text{N}}_2$ and ${\text{O}}_2$ are $0.49\text{ atm ,0}\text{.41 atm ,0}\text{.25 atm and 0}\text{.041 atm}$, respectively.