Question

Use the data provided in a graphical method to determine the order and rate constant of the following reaction:\({\bf{2P}} \to {\bf{Q}} + {\bf{W}}\)

Time (s)	9.0	13.0	18.0	22.0	25.0
(P) (M)	1.077 × 10−3	1.068 × 10−3	1.055 × 10−3	1.046 × 10−3	1.039 × 10−3

Short answer

The order of the reaction is the first order. The rate constant of the reaction is \({\bf{2}}{\bf{.6}} \times {\bf{1}}{{\bf{0}}^{{\bf{ - 3}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

Step-by-step solution

Step 1:  Plotting of graph

The plot of ln(P) vs time is linear. This indicates first-order reaction kinetics.

 Calculation of rate constant 

The constant rate of a first-order reaction is as follows:

\({\bf{K = }}\frac{{{\bf{In}}\frac{{{{{\bf{(P)}}}_{\bf{0}}}}}{{{\bf{(P)}}}}}}{{\bf{t}}}\)

Time	(P)	ln(P)
9	1.077 × 10−3	-6.833
13	1.068 × 10−3	-6.841
18	1.055 × 10−3	-6.854
22	1.046 × 10−3	-6.862
25	1.039 × 10−3	-6.869

Now, the slope of the plot is given as

\(Slope = \frac{{\Delta y}}{{\Delta x}}\)

Where \({\bf{\Delta y}}\) is the change in values of (ln P) at two different times t, and \({\bf{\Delta }}x\) is the corresponding difference in time.

\(Slope = \frac{{( - 6.8540) - ( - 6.841)}}{{18 - 13}} = - 2.6*{10^{ - 3}}{s^{ - 1}}\)

Calculation

Rate= -(slope)

\(\begin{align} &= - ( - 2.6*{10^{ - 3}}{s^{ - 1}}){s^{ - 1}}\\ &= 2.6*{10^{ - 3}}{s^{ - 1}}\end{align}\)

Thus, the order of the reaction is first order with a rate constant of \({\bf{2}}{\bf{.6}} \times {\bf{1}}{{\bf{0}}^{{\bf{ - 3}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)