Chapter 14, Problem 113AP

The first-order rate constant for the decomposition of dimethyl ether:

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\left(g\right)\underset{}{\to }{\text{ CH}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}(g)+CO(g)$

is $\text{3}\text{.2 }\times {\text{ 10}}^{\text{-4}}{\text{ s}}^{\text{-1}}\text{ at 450°C}$. The reaction is carried out in a constant-volume flask. Initially only dimethyl ether is present and the pressure is 0.350 atm. What is the pressure of the system after 8.0 min? Assume ideal behavior.

Interpretation Introduction

Interpretation:

The pressure of the system is 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 by:

$PV=nRT$

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

The formula for conversion of temperature from degree Celsius to Kelvin is represented as:

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

Answer to Problem 113AP

Solution: $0.45atm$

Explanation of Solution

Given information:

Rate constant, $k=3.2\times {10}^{-4}{s}^{-1}$

Temperature, $T={450}^{\text{o}}\text{C}$

Pressure, $P=0.350\text{}\text{atm}$

Time, $t=8.0\min$.

The temperature is $\text{450 }°\text{C}$.

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

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =\left(450+273\right)\\ =723\text{ K}\end{array}$

The standard form of an ideal gas equation is:

$PV=nRT$

Rearrange the above equation for number of moles of hydrogen as follows:

$n=\frac{RT}{PV}$ $n=\frac{PV}{RT}$

Substitute, $0.350\text{ atm}$ for $P$, $\text{723 K}$ for $T$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$, and $V$ for $V$ in the above equation,

$\begin{array}{c}n=\frac{\left(0.350\cancel{\text{atm}}\right)V}{0.0821\text{ L}.\cancel{\text{atm}}\text{/mol}\text{.}\cancel{\text{K}}\left(723\cancel{\text{K}}\right)}\\ =\frac{0.0350\text{V}}{59.358}\\ =0.00589V\\ =5.90\times {10}^{-3}V\end{array}$

According to the Arrhenius equation $\left(K{\text{ = Ae}}^{-Ea/RT}\right)$, the concentration of dimethyl ether is:

$\begin{array}{c}\ln \frac{{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right]}_t}{{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right]}_0}=-kt\\ \frac{{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right]}_t}{{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right]}_0}=e^{-kt}\end{array}$

The concentration of dimethyl ether after $480\text{ s}$ is:

$\begin{array}{c}{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right]}_t=\left(\frac{5.90\times {10}^{-3}\cancel{V}}{\cancel{V}}\right)e^{-\left(3.2\times {10}^{-4}\frac{1}{\cancel{\text{s}}}\right)480\cancel{\text{s}}}\\ =5.06\times {10}^{-3}M\end{array}$ $\begin{array}{c}{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{O}\right]}_t=\left(\frac{5.90\times {10}^{-3}V}{V}\right)e^{-\left(3.2\times {10}^{-4}\frac{1}{\cancel{\text{s}}}\right)480\cancel{\text{s}}}\\ =5.06\times {10}^{-3}M\end{array}$

Thus, after $8.0\min$, the concentration of dimethyl ether is $5.06\times {10}^{-3}M$.

After $8.0\min$, the concentration of dimethyl ether decreased as:

$\left(5.90\times {10}^{-3}-5.06\times {10}^{-3}\right)M=8.4\times {10}^{-4}M$

There are three moles of product formed from each mole of dimethyl ether reacting. So, the concentration of products is:

$\left(3\right)\left(8.4\times {10}^{-4}M\right)=2.5\times {10}^{-3}M$

Molarity is equal to the ratio of number of moles to the volume as follows:

$\frac{n}{V}=M$

The pressure of dimethyl ether after $8.0\min$ is as follows:

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

Substitute $\frac{n}{V}$ for $M$ in the above equation,

$P=MRT$

Substitute $5.0598\times {10}^{-3}M+2.5\times {10}^{-3}M$ for $M$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$, and $\text{723 K}$ for $T$ in the above equation,

$\begin{array}{c}P=\left[\left(5.0598\times {10}^{-3}+2.5\times {10}^{-3}\right)M\right]\left(0.0821\text{ L}\text{.atm}/\text{mol}\text{.}\cancel{\text{K}}\right)\left(723\cancel{\text{K}}\right)\\ =\left(0.00755\cancel{\text{mol}}\text{/}\cancel{\text{L}}\right)\left(59.358\cancel{\text{L}}\text{. atm /}\cancel{\text{mol}}\right)(M=\text{mol / L)}\\ =0.45\text{atm}\end{array}$ $\begin{array}{c}P=\left[\left(5.0598\times {10}^{-3}+2.5\times {10}^{-3}\right)M\right]\left(0.0821\text{ L}\text{.atm}/\text{mol}\text{.}\cancel{\text{K}}\right)\left(723\cancel{\text{K}}\right)\\ =\left(0.00755\right)\left(59.358\right)\\ =0.45\text{atm}\end{array}$

$8.0\min$ $0.45\text{}atm$

Conclusion

The pressure of the system (dimethyl ether) is $0.45atm$.