Rate Law

University:
Eastern Michigan University
Course:
HIST 375 | Modern India
Academic year:

2024
Views:

0

Pages:

2
Author:

Navine Saleh

Determination of a Rate Law from Experimental Data Example 1: Consider the following gas phase reaction: 2NO + 2H2N2 + 2 H2O These data were collected: Experiment 1 2 3 4 [NO]。 (M) [H2]0 (M) A[N2]/At (M/h) 0.60 0.15 0.076 0.60 0.30 0.15 0.60 0.60 0.60 0.30 1.20 1.21 a. Determine the rate law. b. Calculate the value of the rate constant. SOLUTION a. The general form of the rate law is rate must determine the exponents x and y. k[NO]*[H2] To determine the rate law we To determine the value of y, the "order with respect to H2", first find two experiments in which [NO] is held constant. This can be done by any combination of experiments 1, 2, and 3. Comparing the rates of reaction for 1 and 2, notice that the rate of the reaction doubles when the [H2] doubles. This means that the reaction is first order in hydrogen, i.e. y = 1. This is confirmed by comparing the rates of experiments 2 and 3. To determine the order with respect to NO, i.e. to evaluate x, select experiments in which the [H2] is constant while [NO] is allowed to vary. Using experiments 3 and 4, it is clear that the rate quadruples when [NO] doubles. This means that the reaction is second order with respect to NO. The experimental rate law is then Rate = k[NO]2[H2] b. Once the experimental rate law is known, the value of the rate constant can be calculated by inserting a set of experimental data into the rearranged rate law: k = rate of reaction [NO]2[H2] Using the data from the first experiment, k = 1.4 M²h-1 You should get the same, or nearly the same value for k from all trials. Notice that the units for k are consistent with a third order rate law. Always check units to make sure they are consistent with the overall reaction order. Example 2: Consider the following gas phase reaction: 2 NO + O2 2 NO2 These data were collected: Experiment 1 2 3 [NO]. (M) [02] (M) Rate (M/s) 0.0240 0.0350 0.143 0.0150 0.0350 0.0559 0.0240 0.0450 0.184 Determine the rate law. SOLUTION The general form of the rate law is rate = k[NO]*[O2]. To determine the rate law means to determine the numerical value of x and of y. Since there are no cases where the concentration of one component is held constant while the concentration of the other component is simply doubled or tripled, a slightly different approach must be used. Using two trials in which one component concentration is held constant and the other varied, substitute into the rate law for each trial. Then divide one rate expression into the other. Using experiments 1 and 2: rate 1 0.142 M/s k[0.0240]*[0.0350] - -(0.0240) rate 2 0.0559 M/s k[0.0150]*[0.0350] which reduces to 2.56 = (1.60)* which can be solved for x log 2.56 x log 1.60 x = 2.00 Once the order with respect to NO is established, any two sets of data can be used to determine the value of y. Using experiments 1 and 3: rate 3 rate 1 0.143 M/s 0.184 M/s k[0.0240]2[0.0350] k[0.0240]2[0.0450] 0.0350 0.0450, which reduces to 0.777 = (0.777) which can be solved for y log 0.777y log 0.777 y = 1.00 The final rate law is then: rate = k[NO]2[O2] Note that the rate law is NEVER derived from the stoichiometry of the chemical equation. You must have EXPERIMENTAL DATA to establish the relationship between rate and concentration. Once the rate law is known, rates at other concentrations can be predicted.

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