
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 |
- 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.
- 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.
- What is the instantaneous rate, A[B]/Ar, when [BI = 0.750 mol/L?
a)

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

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
Explanation of Solution
The given reaction is:
Based on the stoichiometry of this reaction, the rate can be expressed as:
For the time interval between t = 10 to t = 20 s:
(c)

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:
Therefore, the instantaneous rate is
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Chapter 11 Solutions
Bundle: Chemistry for Engineering Students, Loose-Leaf Version, 4th + OWLv2 with MindTap Reader with Student Solutions Manual, 1 term (6 months) Printed Access Card
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