Chapter 14, Problem 82GQ

We know that the decomposition of SO₂Cl₂ is first-order in SO₂Cl₂,

SO₂Cl₂ → SO₂(g) + Cl₂(g)

with a half-life of 245 minutes at 600 K. If you begin with a partial pressure of SO₂Cl₂ of 25 mm Hg in a 1.0-L flask what is the partial pressure of each reactant and product after 245 minutes? What is the partial pressure of each reactant and product after 12 hours?

Interpretation Introduction

Interpretation:

For the given reaction under given conditions partial pressure of each reactant and product after 245 minutes and after 12 hours should be determined.

Concept introduction:

In order to establish the plausibility of a mechanism, one must compare the rate law of the rate determining step to the experimentally determined rate law.

Rate determining step: In a chemical reaction the rate determining step is the slowest step in which the rate of the reaction depends on the rate of that slowest step.

Rate law: It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Rate constant: The rate constant for a chemical reaction is the proportionality term in the chemical reaction rate law which gives the relationship between the rate and the concentration of the reactant present in the chemical reaction.

Rate order: The order of each reactant in a reaction is represented by the exponential term of the respective reactant present in the rate law and the overall order of the reaction is the sum of all the exponents of all reactants present in the chemical reaction.  The order of the reaction is directly proportional to the concentration of the reactants.

Half-life${\text{t}}_{\text{1/2}}$: The half-life period is defined as the time required by the reactant to reduce its concentration to one half of its initial concentration.

Activation energy: It is defined as the minimum energy required by the reacting species in order to undergo chemical reaction.

Intermediate species: It is the species formed during the middle of the chemical reaction between the reactant and the desired product.

Answer to Problem 82GQ

The partial pressure for the given gas at given time 245 minutes is $\text{12}\text{.5 mm Hg}$ and the partial pressure for $HOF$ at 45 minutes is $3.2mmHg$

Explanation of Solution

Given:

$\begin{array}{l}{\text{Partial pressure of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{ = 25mmHg}\\ \text{Temperature}\text{=}{\text{25}}^{\text{o}}\text{C}\\ \text{Partial pressure after 245 minutes = ?}\\ \text{Partial pressure after 12 hours = ?}\\ \text{Half life = 245 minutes}\\ \text{Volume = 1L}\end{array}$

In order to find the pressure first the rate constant value should be calculated which is determined as follows,

  $\begin{array}{l}{\text{t}}_{\text{1/2}}=\frac{0.693}{k}\\ k=\frac{0.693}{{\text{t}}_{\text{1/2}}}=\frac{0.693}{\text{245 m}}=2.83\times {10}^{-3}{\min }^{-1}\end{array}$

Now, the pressure for the given gas after 245 minutes is calculated as follows,

  $\begin{array}{l}k=\frac{2.303}{t}\log \frac{P_0}{P}\\ 2.83\times {10}^{-3}{\min }^{-1}=\frac{2.303}{245}\log \frac{25mmHg}{P}\\ \log \frac{25mmHg}{P}=\frac{245\times 2.83\times {10}^{-3}{\min }^{-1}}{2.303}\\ \log \frac{25mmHg}{P}=0.3\\ \frac{25mmHg}{P}={10}^{0.3}\\ \frac{P}{25mmHg}=\frac{1}{2}\\ P=\frac{25mmHg}{2}=12.5mmHg\\ \text{Partial}\text{Pressure}{\text{of SO}}_{\text{2}}\text{=Total Pressure}\times \text{mole fraction}\\ =12.5\times \frac{1}{2}=6.25mmHg\\ \text{Partial}\text{Pressure}{\text{of Cl}}_{\text{2}}\text{=Total Pressure}\times \text{mole fraction}\\ =12.5\times \frac{1}{2}=6.25mmHg\\ \text{Partial}\text{Pressure}{\text{of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{=Total Pressure}\times \text{mole fraction}\\ =12.5\times \frac{1}{1}=12.5mmHg\end{array}$

Now, the pressure for the given gas after 12 hours is calculated as follows,

  $\begin{array}{l}k=\frac{2.303}{t}\log \frac{P_0}{P}\\ 2.83\times {10}^{-3}{\min }^{-1}=\frac{2.303}{720}\log \frac{25mmHg}{P}\\ \log \frac{25mmHg}{P}=\frac{720\times 2.83\times {10}^{-3}{\min }^{-1}}{2.303}\\ \log \frac{25mmHg}{P}=0.885\\ \frac{25mmHg}{P}={10}^{0.885}\\ \frac{P}{25mmHg}=\frac{1}{7.7}\\ P=\frac{25mmHg}{7.7}=3.2mmHg\\ \text{Partial}\text{Pressure}{\text{of SO}}_{\text{2}}\text{=Total Pressure}\times \text{mole fraction}\\ =3.2\times \frac{1}{2}=1.6mmHg\\ \text{Partial}\text{Pressure}{\text{of Cl}}_{\text{2}}\text{=Total Pressure}\times \text{mole fraction}\\ =3.2\times \frac{1}{2}=1.6mmHg\\ \text{Partial}\text{Pressure}{\text{of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{=Total Pressure}\times \text{mole fraction}\\ =3.2\times \frac{1}{1}=3.2mmHg\end{array}$

Conclusion

The partial pressure for the given gas at given time 245 minutes was found as $\text{12}\text{.5 mm Hg}$ and the partial pressure for $HOF$ at 45 minutes was found as $3.2mmHg$