Chapter 11, Problem 86AE

Sulfuryl chloride (SO₂Cl₂) decomposes to sulfur dioxide (SO₂) and chlorine (Cl₂) by reaction in the gas phase. The following pressure data were obtained when a sample containing 5.00 × 10⁻² mol sulfuryl chloride was heated to 600. K in a 5.00 × 10⁻¹-L container.

Time (hours):	0.00	1.00	2.00	4.00	8.00	16.00
${\text{P}}_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\left(\text{atm}\right):$	4.93	4.26	3.52	2.53	1.30	0.34

Defining the rate as $-\frac{\Delta \left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}{\Delta t}$,

a. determine the value of the rate constant for the decomposition of sulfuryl chloride at 600. K.

b. what is the half-life of the reaction?

c. what fraction of the sulfuryl chloride remains after 20.0 h?

Interpretation Introduction

Interpretation: The pressure data obtained when a sample containing $5.00\times {10}^{-2}\text{mol}$ sulfuryl chloride was heated to $600\text{K}$ in a $5.00\times {10}^{-1}\text{L}$ container is given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The value of the rate constant $\left(k\right)$ for the decomposition of sulfuryl chloride $\left({\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ at $\text{600}\text{K}$

Answer to Problem 86AE

Answer The value of rate constant is $\underset{\_}{1.68\times 10^{-1}{s}^{-1}}$.

Explanation of Solution

Explanation

Given

Temperature is $\text{600}\text{K}$.

Given table,

Time (hours)	$\text{0}\text{.00}$	$\text{1}\text{.00}$	$\text{2}\text{.00}$	$\text{4}\text{.00}$	$\text{8}\text{.00}$	$\text{16}\text{.00}$
$P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\left(\text{atm}\right)$	$\text{4}\text{.93}$	$\text{4}\text{.26}$	$\text{3}\text{.52}$	$\text{2}\text{.53}$	$\text{1}\text{.30}$	$\text{0}\text{.34}$

The formula of ideal gas law is,

$\begin{array}{c}PV=nRT\\ \frac{n}{V}=\frac{P}{RT}\\ M=\frac{P}{RT}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.
• $M$ is the concentration.

The concentration of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is calculated for each given pressure using the above equation. The value of concentration and $\ln \left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]$ is,

Pressure	Concentration of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ $\left(M=\frac{P}{RT}\right)$	$\ln \left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]$
$\text{4}\text{.93}$	$\text{0}\text{.100079}$	$-\text{2}\text{.3}$
$\text{4}\text{.26}$	$\text{0}\text{.0865}$	$-\text{2}\text{.45}$
$\text{3}\text{.52}$	$\text{0}\text{.0715}$	$-\text{2}\text{.64}$
$\text{2}\text{.53}$	$\text{0}\text{.0513}$	$-\text{2}\text{.97}$
$\text{1}\text{.3}$	$\text{0}\text{.026}$	$-\text{3}\text{.65}$
$\text{0}\text{.34}$	$\text{0}\text{.0069}$	$-\text{4}\text{.98}$

The graph is plotted for $\text{ln}\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]$ versus time is,

Since, the obtained graph is a straight line, hence, the order of the reaction is first. The slope of graph is $-\text{0}\text{.168}$.

The integral rate law equation of first order reaction is,

$\text{ln}\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]=-kt+\text{ln}{\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}_{\text{0}}$ (1)

Where,

• $k$ is the rate constant.
• ${\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}_{\text{0}}$ is the initial concentration of reactant.
• $t$ is the time.

The equation (1) is similar to the equation of a straight line, that is,

$y=mx+c$ (2)

Where,

• $y$ is the y-intercept.
• $x$ is the x-intercept.
• $m$ is the slope.
• $c$ is a constant.

Compare equation (1) and (2).

$m=-k$

Substitute the value of slope in the above equation.

$\begin{array}{c}m=-k\\ k=-\left(-\text{0}\text{.168}{\text{s}}^{-1}\right)\\ =\underset{\_}{1.68\times 10^{-1}{s}^{-1}}\end{array}$

Interpretation Introduction

Interpretation: The pressure data obtained when a sample containing $5.00\times {10}^{-2}\text{mol}$ sulfuryl chloride was heated to $600\text{K}$ in a $5.00\times {10}^{-1}\text{L}$ container is given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The half life of the given reaction.

Answer to Problem 86AE

Answer

The half life of the given reaction is $\underset{\_}{4.12s}$.

Explanation of Solution

Explanation

The value of rate constant is $\text{1}\text{.68}\times {\text{10}}^{-\text{1}}{\text{s}}^{-\text{1}}$.

Formula

The half-life is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

Where,

• $t_{\frac{\text{1}}{\text{2}}}$ is half life.
• $k$ is rate constant.

Substitute the values of $k$ in the above equation.

$\begin{array}{c}{t}_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}\\ =\frac{\text{0}\text{.693}}{\text{1}\text{.68}\times {\text{10}}^{-\text{1}}{\text{s}}^{-\text{1}}}\\ =\underset{\_}{4.12s}\end{array}$

Interpretation Introduction

Interpretation: The pressure data obtained when a sample containing $5.00\times {10}^{-2}\text{mol}$ sulfuryl chloride was heated to $600\text{K}$ in a $5.00\times {10}^{-1}\text{L}$ container is given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The fraction of sulfuryl chloride $\left({\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ left after $\text{20}\text{hours}$.

Answer to Problem 86AE

Answer

The fraction of sulfuryl chloride $\left({\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ left after $\text{20}\text{hours}$ is $\underset{\_}{7.4\times {10}^{-5}mol/L}$.

Explanation of Solution

Explanation

Given

The initial partial pressure of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at $\text{4}\text{.93}\text{atm}$ is $\text{0}\text{.100079}\text{mol/L}$.

The value of rate constant is $\text{1}\text{.68}\times {\text{10}}^{-\text{1}}{\text{s}}^{-\text{1}}$.

Time is $\text{20}\text{hours}$.

The conversion of hours $\left(\text{hr}\right)$ into seconds $\left(\text{s}\right)$ is done as,

$\text{1}\text{hr}=\text{3600}\text{s}$

Hence,

The conversion of $\text{20}\text{hr}$ into seconds is,

$\begin{array}{c}\text{20}\text{hr}=\left(\text{20}\times \text{3600}\right)\text{s}\\ =\text{72000}\text{s}\end{array}$

Formula

The rate constant of first order reaction is,

$k=\frac{\text{2}\text{.303}}{t}\log \left(\frac{{\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}_{\text{0}}}{\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}\right)$

Where,

• $k^{\text{'}}$ is the rate constant.
• ${\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}_{\text{0}}$ is the initial partial pressure of reactant.
• $\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]$ is the fraction of reactant left.
• $t$ is the time.

Substitute the values of ${\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}_{\text{0}},k$ and $t$ in the above equation.

$\begin{array}{c}k=\frac{\text{2}\text{.303}}{t}\log \left(\frac{{\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}_{\text{0}}}{\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}\right)\\ \text{0}\text{.168}{\text{s}}^{-\text{1}}=\frac{\text{2}\text{.303}}{\text{72000}\text{s}}\log \left(\frac{\text{0}\text{.100079}}{\left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]}\right)\\ \left[P_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\right]=\underset{\_}{7.4\times {10}^{-5}mol/L}\end{array}$

Conclusion

Conclusion

The plot of $\ln \left[\text{A}\right]\text{vs}\text{time}$ graph is a straight line. It means that the given reaction is the first order $\left(a=\text{1}\right)$ reaction.

The integral rate law equation of first order reaction is,

$\ln \left[\text{A}\right]=-kt+\ln {\left[\text{A}\right]}_{\text{0}}$