Chapter 14, Problem 22QP

The following gas-phase reaction was studied at $\text{290°C}$ by observing the change in pressure as a function of time in a constant-volume vessel:

${\text{ClCO}}_{\text{2}}{\text{CCl}}_{\text{3}}\left(g\right)\to 2{\text{COCl}}_{\text{2}}\left(g\right)$

Determine the order of the reaction and the rate constant based on the following data.

Time (s)	P(mmHg)
0	15.76
181	18.88
513	22.79
1164	27.08
where P is the total pressure.

Interpretation Introduction

Interpretation:

The order of the reaction and the rate constant, based on the given data, areto be determined.

Concept introduction:

The power of the concentration of reactants is called the order of the reaction.

According to the Arrheniusequation, the rate that is dependent on a given temperature is called the rate constant.

Answer to Problem 22QP

Solution: The order of the reaction is first order and the rate constant is $1.08\times {10}^{-3}$.

Explanation of Solution

Given information: The following gas phase reaction was studied at a temperature of $290°\text{C}$ by observing the change in pressure as a function of time in a constant-volume vessel.

The reaction is as follows:

${\text{ClCO}}_{\text{2}}{\text{CCl}}_3\left(g\right)\to 2{\text{COCl}}_{\text{2}}\left(g\right)$

The given data is as follows:

$\begin{array}{c}\underset{\_}{t\left(s\right)P\left(\text{mmHg}\right)}\\ 0\text{ 15}\text{.76 }\\ 181\text{ 18}\text{.88 }\\ 513\text{ 22}\text{.79 }\\ 1164\text{ 27}\text{.08 }\end{array}$

Consider $P_0$ as the pressure of the reactant in the above reaction when the time $\left(t\right)$ is $0$ and $x$ is considered as the pressure decrease after $t$. In the abovereaction, if ${\text{ClCO}}_{\text{2}}{\text{CCl}}_3$ loses one atmosphere, two atmospheres of ${\text{COCl}}_{\text{2}}$ areformed.

The reaction is as follows:

${\text{ClCO}}_{\text{2}}{\text{CCl}}_3\left(g\right)\to 2{\text{COCl}}_{\text{2}}\left(g\right)$

The pressure data is represented below.

$\begin{array}{l}\underset{\_}{\text{Time }\left[{\text{ClCO}}_{\text{2}}{\text{CCl}}_3\right]\left[{\text{COCl}}_{\text{2}}\right]}\\ t=0\text{}{P}_0\text{ 0}\\ t=t{P}_0-x\text{ 2}x\end{array}$

Therefore, the pressure increases, and the change in pressure ($\Delta P$) is as follows:

$\left(2x-x\right)=x$

The data is represented as follows:

$\begin{array}{c}\underset{\_}{t\left(s\right)P\left(\text{mmHg}\right)\frac{1}{P_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}}\Delta P=x{P}_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}\text{ ln }{P}_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}}\\ 0\text{ 15}\text{.76 0}\text{.00 15}\text{.76 2}\text{.757 0}\text{.0635}\\ 181\text{ 18}\text{.88 3}\text{.12 12}\text{.64 2}\text{.537 0}\text{.0791}\\ 513\text{ 22}\text{.79 7}\text{.03 8}\text{.73 2}\text{.167 0}\text{.115}\\ 1164\text{ 27}\text{.08 11}\text{.32 4}\text{.44 1}\text{.491 0}\text{.225}\end{array}$

The plot of $\ln P_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}$ vs. $t$ will be in a linear form, if the reaction is of the first order. The plot of $\frac{1}{P_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}}$ vs. $t$ will be linear, if the reaction is of the second order.

The plot of $\ln P_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}$ vs. $t$ is as follows:

The plot of $\frac{1}{P_{{\text{ClCO}}_{\text{2}}{\text{CCl}}_3}}$ vs. $t$ is as follows:

The rate constant for the first-order reaction is as follows:

$\begin{array}{c}\text{slope}=\frac{y_2-y_1}{x_2-x_1}\\ =\left(\frac{2.757-1.491}{1164}\right)\\ =\left(\frac{1.266}{1164}\right)\\ =0.00108\text{ or }1.08\times {10}^{-3}\end{array}$

The reaction is a first-order reaction, as can be interpreted from the graph. The slope for a first-order reaction is equal to the rate constant $\left(-k\right)$.

Conclusion

The order of reaction is first order and rate constant is $1.08\times {10}^{-3}$.