Question 14.4.5: Use the graphical method to determine reaction order and rat......
Use the graphical method to determine reaction order and rate constant.
Concentration versus time data have been collected for the decomposition of H_{2}O_{2} at 300 K.
H_{2}O_{2}(aq) → H_{2}O(ℓ) + ½ O_{2}(g)
Use these data to determine the order of the reaction with respect to H_{2}O_{2} and the rate constant for the reaction at this temperature.
| \pmb{[H_{2}O_{2}]} (mol/L) | Time (min) |
| 0.0200 | 0 |
| 0.0118 | 500. |
| 0.00693 | 1000. |
| 0.00408 | 1500. |
| 0.00240 | 2000. |
| 0.00141 | 2500. |
You are asked to determine the order of a reaction with respect to a reactant.
You are given the balanced equation for the reaction and data showing reactant concentration as a function of time.
Use the concentration–time data to calculate ln[H_{2}O_{2}] and 1/[H_{2}O_{2}], and use graphing software to create three plots: [H_{2}O_{2}] versus time, ln[H_{2}O_{2}] versus time, and 1/[H_{2}O_{2}] versus time.
The data fall on a straight line only in the ln[H_{2}O_{2}] versus time plot. Thus, the reaction is first order with respect to H_{2}O_{2}. The rate law for the reaction is
rate = k[H_{2}O_{2}]
The value of the rate constant is found from the slope of the ln[H_{2}O_{2}] versus time plot. Using the graphing software, the equation for the straight line that passes through the data points in this plot is y = −0.00106x − 3.912.
For a first-order reaction, when the concentration–time data are plotted as a natural log of concentration versus time, the slope of the line is equal to −k. For this reaction, k = −(−0.00106 min^{-1}) = 0.00106 min^{-1} at 300 K
