Chapter 15, Problem 108AE

(a)

Interpretation Introduction

Interpretation:

The P_total at $t=80\text{ s}$ should be predicted.

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 108AE

  ${\text{P}}_{total}=370\text{ torr}$

Explanation of Solution

  ${\text{C}}_2{\text{H}}_5\text{OH }\to {\text{ C}}_2{\text{H}}_4{\text{ + H}}_2\text{O}$

The order of the reaction should be determined first.

Initial pressure of C₂H₅OH = ${\text{P}}_0=250\text{ torr}$

If x is the pressure of C₂H₅OH reacts, then at any time, t

  ${\text{P}}_{C_2{H}_{5}OH}=250-x$

  $\begin{array}{l}{\text{P}}_{C_2{H}_4}={\text{P}}_{H_{2}O}=x\\ {\text{P}}_{total}=250-x+x+x=250+x\end{array}$

  $x={\text{P}}_{total}-250$

t (s)	PTotal (torr)	x (torr)	Pethanol (torr)
0	250	0	250
10	265	15	235
20	280	30	220
30	295	45	205
40	310	60	190
50	325	75	175

  

P_ethanol versus time plot is linear. So, decomposition of ethanol follows zeroth order kinetics with respect to ethanol.

According to the equation on chart, P_total can be calculated as follows.

  $\begin{array}{l}y=-1.5x+250\\ {\text{P}}_{ethanol}=-1.5\times 80\text{ s + 250 = 130 torr}\end{array}$

  $\begin{array}{l}{\text{P}}_{ethanol}=250-x\\ 130\text{ torr = 250 torr }-x\\ x=120\text{ torr}\end{array}$

  $\begin{array}{l}{\text{P}}_{total}=250\text{ +}x=250+120\\ {\text{P}}_{total}=370\text{ torr}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of the rate constant and the units 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 108AE

  $k=1.5\text{ torr/s}$

Explanation of Solution

P_ethanol versus time plot is linear. So, decomposition of ethanol follows zeroth order kinetics with respect to ethanol.

Integrated rate law for the reaction; ${\text{P}}_{{\text{C}}_2{\text{H}}_5\text{OH}}={\left({\text{P}}_{{\text{C}}_2{\text{H}}_5\text{OH}}\right)}_0-kt$

So, the slope of the graph gives the value of the rate constant.

  $k=1.5\text{ torr/s}$

(c)

Interpretation Introduction

Interpretation:

The order of the reaction 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 108AE

Zeroth order

Explanation of Solution

P_ethanol versus time plot is linear. So, decomposition of ethanol follows zeroth order kinetics with respect to ethanol.

Integrated rate law for the reaction; ${\text{P}}_{{\text{C}}_2{\text{H}}_5\text{OH}}={\left({\text{P}}_{{\text{C}}_2{\text{H}}_5\text{OH}}\right)}_0-kt$

(d)

Interpretation Introduction

Interpretation:

P_total at $t=300\text{ s}$ should be calculated.

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 108AE

  ${\text{P}}_{total}=500\text{ torr}$

Explanation of Solution

According to the equation on chart, P_total can be calculated as follows.

  $\begin{array}{l}y=-1.5x+250\\ {\text{P}}_{ethanol}=-1.5\times 300\text{ s + 250 = }-200\text{ torr}\end{array}$

  $\begin{array}{l}{\text{P}}_{ethanol}=250-x\\ -200\text{ torr = 250 torr }-x\\ x=450\text{ torr}\end{array}$

  $\begin{array}{l}{\text{P}}_{total}=250\text{ + }x=250+450\\ {\text{P}}_{total}=700\text{ torr}\end{array}$

But this is an impossible answer. Because the maximum pressure that can be after the completion of reaction is 500 torr.

When all the ethanol is decomposed

  ${\text{P}}_{total}={\text{P}}_{C_2{H}_4}+{\text{P}}_{H_{2}O}=250+250=500\text{ torr}$

Therefore, at $t=300\text{ s}$ , ${\text{P}}_{total}=500\text{ torr}$