Chapter 11, Problem 11.20PAE

Experimental data are listed here for the reaction B:

Time (s)	IB] (mol/L)
0.00	0.000
10.0	0.326
20.0	0.572
30.0	0.750
40.0	0.890

1. Prepare a graph from these data, connect the points with a smooth line, and calculate the rate of change of [B] for each 10-s interval from 0.0 to 40.0 s. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result.
2. How is the rate of change of [AJ related to the rate of change of [B] in each time interval? Calculate the rate of change of [AJ for the time interval from 10.0 to 20.0 s.
3. What is the instantaneous rate, A[B]/Ar, when [BI = 0.750 mol/L?

Interpretation Introduction

Interpretation:

A graph based on the given data must be plotted and the rate of change of [B] for every 10 s interval must be calculated.

Concept Introduction:

• Chemical reactions proceed at a certain rate which is represented in terms of the change in concentration over a certain period of time
• The rate can be expressed either in terms of a decrease in concentration of the reactants or an increase in the concentration of products.

Answer to Problem 11.20PAE

Solution:

The plot is depicted below.

The rate of change of [B] decreases from one-time interval to the next.

Explanation of Solution

The given reaction is:

$\text{A }\to \text{ 2B}$

A plot of concentration of [B] vs time based on the given data is shown below:

$\begin{array}{l}\text{ Rate = }\frac{\Delta \text{[B]}}{\Delta \text{t}}\text{ = }\frac{{\text{[B]}}_{{\text{t}}_{\text{2}}}{\text{ - [B]}}_{{\text{t}}_{\text{1}}}}{{\text{t}}_2{\text{ - t}}_{\text{1}}}\\ {\text{ 1) From t}}_{\text{1}}{\text{ = 0 to t}}_{\text{2}}\text{ = 10 s }\\ \text{ Rate = }\frac{[0.326\text{ - 0}\text{.000] mol/L}}{[10.0\text{ - 0}\text{.0] s }}\text{ = 0}\text{.033 mol/L-s}\\ {\text{2) From t}}_{\text{1}}{\text{ = 10 to t}}_{\text{2}}\text{ = 20 s }\\ \text{ Rate = }\frac{[0.572\text{ - 0}\text{.326] mol/L}}{[20.0\text{ - 10}\text{.0] s }}\text{ = 0}\text{.025 mol/L-s}\\ {\text{3) From t}}_{\text{1}}{\text{ = 20 to t}}_{\text{2}}\text{ = 30 s }\\ \text{ Rate = }\frac{[0.750\text{ - 0}\text{.572] mol/L}}{[30.0\text{ - 20}\text{.0] s }}\text{ = 0}\text{.018 mol/L-s}\\ {\text{4) From t}}_{\text{1}}{\text{ = 30 to t}}_{\text{2}}\text{ = 40 s }\\ \text{ Rate = }\frac{[0.890\text{ - 0}\text{.750] mol/L}}{[40.0\text{ - 30}\text{.0] s }}\text{ = 0}\text{.014 mol/L-s}\end{array}$

As per the above calculations, the rate of change decreases from one-time interval to the next. This is because as time increases the concentration of reactants decreases as a result the rate of formation of the products will also decrease.

Interpretation Introduction

Interpretation:

The relation between the rate of change of [A] and the rate of change of [B] must be explained. The rate of change of [A] for the time interval from 10.0 to 20.0 s should be calculated.

Concept Introduction:

• Chemical reactions proceed at a certain rate which is represented in terms of the change in concentration over a certain period of time
• The rate can be expressed either in terms of a decrease in concentration of the reactants or an increase in the concentration of products.

Answer to Problem 11.20PAE

Solution:

The rate of change of A is half that of B and for time interval 10 to 20 s rate of change of [A] is $0.013\text{ mol/L-s}$.

Explanation of Solution

The given reaction is:

$\text{A }\to \text{ 2B}$

Based on the stoichiometry of this reaction, the rate can be expressed as:

$\text{Rate = }-\frac{\text{Δ[A]}}{\text{Δt}}\text{ = }\frac{1}{2}\frac{\text{Δ[B]}}{\text{Δt}}$

For the time interval between t = 10 to t = 20 s:

$\begin{array}{l}\frac{\Delta \text{[B]}}{\Delta \text{t}}\text{ = }\frac{{\text{[B]}}_{{\text{t}}_{\text{2}}}{\text{ - [B]}}_{{\text{t}}_{\text{1}}}}{{\text{t}}_2{\text{ - t}}_{\text{1}}}\text{ = }\frac{[0.572\text{ - 0}\text{.326] mol/L}}{[20.0\text{ - 10}\text{.0] s }}\text{ = 0}\text{.025 mol/L-s}\\ \text{Based on equation (1):}\\ \frac{\Delta \text{[A] }}{\Delta \text{t}}\text{ = }\frac{1}{2}\times 0.025=0.013\text{ mol/L-s}\end{array}$

Interpretation Introduction

Interpretation:

The instantaneous rate must be calculated when [B] = 0.750 mol/L

Concept Introduction:

• Chemical reactions proceed at a certain rate which is represented in terms of the change in concentration over a certain period of time
• The rate can be expressed either in terms of a decrease in concentration of the reactants or an increase in the concentration of products.
• Average rate of a reaction can be defined as the difference in the concentrations measured at two different times whereas, instantaneous rate can be defined as the rate of a reaction at a particular instant in time.

Answer to Problem 11.20PAE

Solution: Rate = 0.200 mol/L-s

Explanation of Solution

Instantaneous rate can be deduced by drawing a tangent at the point of the curve that corresponds to a particular instant. The slope of the tangent gives the instantaneous rate. In this case the value of [B] = 0.075 mol/L corresponds to a time t = 30 sec. The tangent and the slope are depicted in the plot shown below:

$\begin{array}{c}\text{Instantaneous Rate = }\frac{\text{ΔY}}{\text{ΔX}}\\ \text{= }\frac{[0.800-0.600]\text{ mol/L}}{[33.0-32.0]\text{ s}}\\ \text{= 0}\text{.200 mol/L-s}\end{array}$

Therefore, the instantaneous rate is $\text{0}\text{.200 mol/L-s}$.