Chapter 15, Problem 106AE

(a)

Interpretation Introduction

Interpretation:

The value of the rate constant for the decomposition of sulfuryl chloride at 600 K should be determined.

Concept Introduction:

Integrated rate laws for zero, first and second order reactions are,

Zeroth order: ${\left[\text{A}\right]}_t={\left[\text{A}\right]}_0-kt$

First order: $\ln {\left[\text{A}\right]}_t=\ln {\left[\text{A}\right]}_0-kt$

Second order: $\frac{1}{{\left[\text{A}\right]}_t}=kt+\frac{1}{{\left[\text{A}\right]}_0}$

Answer to Problem 106AE

  $k=0.168{\text{ hr}}^{-1}$

Explanation of Solution

  ${\text{SO}}_2{\text{Cl}}_2\text{(g) }\to {\text{ SO}}_2{\text{(g) + Cl}}_2\text{(g)}$

  ${\text{P}}_0={\text{ initial pressure of SO}}_2{\text{Cl}}_2$

If $x={\text{P}}_{{\text{SO}}_2}$ , then $x={\text{P}}_{{\text{SO}}_2}={\text{P}}_{{\text{Cl}}_2}$ and ${\text{P}}_{{\text{SO}}_2{\text{Cl}}_2}={\text{P}}_0-x$

  $\begin{array}{l}{\text{P}}_{\text{total}}{\text{= P}}_{{\text{SO}}_2{\text{Cl}}_2}+{\text{P}}_{{\text{SO}}_2}+{\text{P}}_{{\text{Cl}}_2}\\ {\text{ = P}}_0-x+x+x\\ {\text{P}}_{\text{total}}={\text{P}}_0+x\\ {\text{P}}_{\text{total}}-{\text{P}}_0=x\end{array}$

Time(hour)	0.00	1.00	2.00	4.00	8.00	16.00
Ptotal(atm)	4.93	5.60	6.34	7.33	8.56	9.52
PSO2Cl2	4.93	4.26	3.52	2.53	1.30	0.34
Ln PSO2Cl2	1.595	1.449	1.258	0.928	0.262	-1.08

  

Pressure of gas is directly proportional to concentration.

The graph of ln P_SO2Cl2 versus time is linear. So, the reaction of decomposition of SO₂Cl₂ is first order.

Integrated rate law for the reaction is, $\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_t=\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_0-kt$

The slope of the graph gives the value for rate constant.

  $k=0.168{\text{ hr}}^{-1}$

(b)

Interpretation Introduction

Interpretation:

The half-life of the reaction should be calculated.

Concept Introduction:

Half-life for first order reaction can be calculated by,

  $t_{1/2}=\frac{\ln 2}{k}$

Answer to Problem 106AE

  $t_{1/2}=4.13\text{ h}$

Explanation of Solution

  $\begin{array}{l}{t}_{1/2}=\frac{\ln 2}{k}\\ \text{ = }\frac{0.693}{0.168{\text{ hr}}^{-1}}\\ t_{1/2}=4.13\text{ h}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The pressure in the vessel after 0.500 h and after 12.0 h should be calculated.

Concept Introduction:

Integrated rate law for the first order reaction;

  $\ln {\left[\text{A}\right]}_t=\ln {\left[\text{A}\right]}_0-kt$

[A]_t − concentration of A at time t

[A]₀ − initial concentration of A

k − rate constant

t − time

Answer to Problem 106AE

After 0.500 h = 5.33 atm

After 12.00 h = 9.20 atm

Explanation of Solution

After 0.500 h;

  $\begin{array}{l}\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_t=\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_0-kt\\ \text{ = ln(4}\text{.93) - 0}{\text{.168 hr}}^{-1}\times 0.500\text{ h}\\ \text{ = -0}\text{.0840}\\ {\text{P}}_{S{O}_{2}C{l}_2}=4.53\text{ atm}\end{array}$

  ${\text{P}}_{C{l}_2}={\text{P}}_{S{O}_2}=4.93\text{ atm - 4}\text{.53 atm = 0}\text{.40 atm}$

  $\begin{array}{l}{\text{P}}_{total}{\text{=P}}_{S{O}_{2}C{l}_2}{\text{+P}}_{C{l}_2}{\text{+P}}_{S{O}_2}\\ =4.53\text{ atm + 0}\text{.40 atm + 0}\text{.40 atm}\\ \text{ = 5}\text{.33 atm}\end{array}$

After 12.00 h;

  $\begin{array}{l}\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_t=\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_0-kt\\ \text{ = ln(4}\text{.93) - 0}{\text{.168 hr}}^{-1}\times 12.00\text{ h}\\ \text{ = -0}\text{.42}\\ {\text{P}}_{S{O}_{2}C{l}_2}=0.66\text{ atm}\end{array}$

  ${\text{P}}_{C{l}_2}={\text{P}}_{S{O}_2}=4.93\text{ atm - 0}\text{.66 atm = 4}\text{.27 atm}$

  $\begin{array}{l}{\text{P}}_{total}{\text{=P}}_{S{O}_{2}C{l}_2}{\text{+P}}_{C{l}_2}{\text{+P}}_{S{O}_2}\\ =0.66\text{ atm + 4}\text{.27 atm + 4}\text{.27 atm}\\ \text{ = 9}\text{.20 atm}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The fraction of the sulfuryl chloride remains after 20.0 h should be calculated.

Concept Introduction:

Integrated rate law for the first order reaction;

  $\ln {\left[\text{A}\right]}_t=\ln {\left[\text{A}\right]}_0-kt$

[A]_t − concentration of A at time t

[A]₀ − initial concentration of A

k − rate constant

t − time

Answer to Problem 106AE

Fraction of SO₂Cl₂ left = $0.0347$

Explanation of Solution

  $\begin{array}{l}\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_t=\ln {\left[{\text{P}}_{S{O}_{2}C{l}_2}\right]}_0-kt\\ \ln \left(\frac{{\text{P}}_{S{O}_{2}C{l}_2}}{{\left({\text{P}}_{S{O}_{2}C{l}_2}\right)}_0}\right)=-0.168{\text{ h}}^{-1}\times 20.0\text{ h}\\ \text{ = }-3.36\\ \frac{{\text{P}}_{S{O}_{2}C{l}_2}}{{\left({\text{P}}_{S{O}_{2}C{l}_2}\right)}_0}=3.47\times {10}^{-2}\end{array}$

Fraction of SO₂Cl₂ left = $0.0347$