Chapter 10, Problem 107AP

Interpretation Introduction

Interpretation:

The gasesthat are left after the reaction are to be identified and their partial pressures are to be determined.

Concept Introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown below:

$P=R\frac{nT}{V}.$

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

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

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

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

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

Here, $n_i$ is the number of moles of the 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

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

Answer to Problem 107AP

Solution: The gases left after the reaction are ${\text{NO}}_2$ and ${\text{O}}_2$, the partial pressure for ${\text{NO}}_2$ is $0.333\text{ atm,}$ and for ${\text{O}}_2$ is $0.166\text{ atm}$.

Explanation of Solution

Given information: The reaction of nitric oxide with molecular oxygen at $25°\text{C}$ is as follows:

$\text{2NO}\left(s\right)+{\text{O}}_2\left(g\right) \to 2{\text{NO}}_2\left(g\right).$

The volume and pressure of NO are $4\text{ L}$ and $0.5\text{ atm}$, respectively, and the volume and pressure of ${\text{O}}_2$ are $2\text{ L}$ and $1\text{ atm}$, respectively.

To determine the number of moles of $\text{NO}$, the ideal gas equation can be rewritten as:

$n_{\text{NO}}=\frac{P{V}_{\text{NO}}}{RT}.$

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

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

The absolute value of $T$ is:

$\begin{array}{c}T\left(\text{in K}\right)=\left[25°\text{C}+273.15\right]\text{ K}\\ =298.15\text{ K}\text{.}\end{array}$

Substitute the values $4\text{ L}$ for $V$, $298.15\text{ K}$ for $T$, $0.5\text{ atm}$ for $P$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the rewritten ideal gas equation as follows:

$\begin{array}{c}{n}_{\text{NO}}=\frac{\left(0.5\cancel{\text{atm}}\right)\left(4\cancel{\text{L}}\right)}{\left(0.08206\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(298.15\cancel{\text{K}}\right)}\\ =\frac{2}{24.47}\text{ mol}\\ =0.0817\text{ mol}\text{.}\end{array}$

To determine the number of moles of ${\text{O}}_2$, the ideal gas equation can be rewritten as follows:

$n_{{\text{O}}_2}=\frac{P{V}_{{\text{O}}_2}}{RT}.$

Substitute the values $2\text{ L}$ for $V$, $298.15\text{ K}$ for $T$, $1\text{ atm}$ for $P$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation as follows:

$\begin{array}{c}{n}_{{\text{O}}_2}=\frac{\left(1\cancel{\text{atm}}\right)\left(2\text{ L}\right)}{\left(0.08206\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(298.15\cancel{\text{K}}\right)}\\ =\frac{2}{24.47}\text{ mol}\\ =0.0817\text{ mol}\text{.}\end{array}$

Now, in the reaction, two moles of $\text{NO}$ arerequired to react with one mole of ${\text{O}}_2$. So, in the reaction, $\text{NO}$ is the limiting reagent as they both have equal number of moles.

As $\text{NO}$ is the limiting reagent in this reaction and is consumed completely in the reaction, the gases left after the reaction are ${\text{NO}}_2$ and ${\text{O}}_2$.

The moles of ${\text{NO}}_2$ produced are

$\begin{array}{c}{n}_{{\text{NO}}_2}=n_{\text{NO}}\\ =0.0817\text{ mol}\text{.}\end{array}$

So, the quantity of ${\text{O}}_2$ spent in the reaction is

$\begin{array}{l}=\frac{0.0817\text{ mol}}{2}\\ =0.04085\text{ mol}\text{.}\end{array}$

Now, the quantity of ${\text{O}}_2$ remaining in the reaction is

$\begin{array}{l}=0.0817\text{ mol}-0.04085\text{ mol}\\ =0.04085\text{ mol}\text{.}\end{array}$

The total volume of the apparatus is

$\begin{array}{c}{V}_T=2\text{ L}+4\text{ L}\\ =6\text{ L}\text{.}\end{array}$

Now, the partial pressure is calculated as follows:

For ${\text{NO}}_2$:

$P_{{\text{NO}}_2}=\frac{n_{{\text{NO}}_2}RT}{V_T}.$

Substitute the values $6\text{ L}$ for $V_T$, $298.15\text{ K}$ for $T$, $0.0817\text{ mol}$ for $n$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation as follows:

$\begin{array}{c}{P}_{{\text{NO}}_2}=\frac{\left(0.0817\cancel{\text{mol}}\right)\left(0.08206\cancel{\text{L}}\text{.atm}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(298.15\cancel{\text{K}}\right)}{\left(6\cancel{\text{L}}\right)}\\ =\frac{1.999}{6}\text{ atm}\\ =0.333\text{ atm}\text{.}\end{array}$

For ${\text{O}}_2$:

$P_{{\text{O}}_2}=\frac{n_{{\text{O}}_2}RT}{V_T}.$

Substitute the values $6\text{ L}$ for $V_T$, $298.15\text{ K}$ for $T$, $0.04085\text{ mol}$ for $n$, and $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$ in the above equation as follows:

$\begin{array}{c}{P}_{{\text{O}}_2}=\frac{\left(0.04085\cancel{\text{mol}}\right)\left(0.08206\cancel{\text{L}}\text{.atm}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(298.15\text{ K}\right)}{\left(6\cancel{\text{L}}\right)}\\ =\frac{0.999}{6}\text{ atm}\\ =0.166\text{ atm}\text{.}\end{array}$

Conclusion

The gases left after the reaction are ${\text{NO}}_2$ and ${\text{O}}_2$, thepartial pressure for ${\text{NO}}_2$ is $0.333\text{ atm,}$ and for ${\text{O}}_2$ is $0.166\text{ atm}$.