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Questions
Describe the graphical representation of first order reaction.
Describe the graphical method for the determination of the order of a first order reaction.
How would you represent a first order reaction graphically?
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Solution 1
The rate constant for first order reaction is,
kt = ln `([A_0])/([A])`
kt = In [A0] – In [A]
In [A] = In [A0] – kt
y = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time intervals ‘t’, a plot of ln [A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated.
A plot of ln [A] vs t for a first-order reaction, \[\ce{A -> product}\] with initial concentration of [A] = 1.00 M and k = 2.5 × 10−2 min−1
Solution 2
A first order reaction can be represented graphically in several ways. Some important graphical representations and their characteristics are as follows.
Reaction rate vs concentration plot: The rate law for a first-order reaction is given by
Rate = k[A]
A plot of reaction rate vs concentration of reactant for a first order reaction.
The equation y = mx indicates a straight line that passes through the origin. Plotting reaction rates against reactant concentration yields a straight line that passes through the origin, as illustrated in the picture above. The slope of the line corresponds to the rate constant of the reaction.
\[\ce{Slope of the line = \frac{CB}{AC} = k}\]
Concentration vs. time plot: We know that
[A] = [A]0 e−kt
Exponential decay of the conc. of reactant as a function of time for a first order reaction.
The equation shows that a first-order process is exponential, with the concentration of reactant decreasing exponentially over time. Graphing reactant concentrations over time yields an exponential decay curve, as illustrated in the image above.
log10 [A] vs time plot: We know that
\[\ce{k = \frac{2.303}{t} log_10 \frac{[A]_0}{[A]}}\]
This equation can be written as
\[\ce{log_10 [A] = - \frac{kt}{2.303} + log_10[A]_0}\]
The above equation is of the type y = mx + c and represents a straight line. Therefore, on plotting log10 [A] against t, a straight line (as shown in the figure above) is obtained. The slope of the line is equal to \[\ce{\frac{-k}{2.303}}\] while the intercept of the line on log10 [A] axis is equal to log10 [A]0. Thus,
\[\ce{Slope of the line = \frac{AC}{BC} = - \frac{k}{2.303}}\]
Intercept on log10 [A] axis = log10 [A]0
Thus, the value of k can be obtained from the slope of the line.
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