Question

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?

Answer 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

Answer 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.