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Chapter 11: Problem 95

The reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow\) products \((a q)\) was studied, and the following data were obtained: What is the order of the reaction with respect to A? What is the order of the reaction with respect to B? What is the value of the rate constant for the reaction?

Short Answer

Expert verified
The reaction is first-order with respect to A and second-order with respect to B. The rate constant for the reaction is \(k = 250 \text{ M}^2/\text{s}\).

Step by step solution

01

Write the Rate Law

The Rate Law for a reaction is given by the following formula: \[rate = k[A]^m[B]^n\] where k is the rate constant, [A] and [B] are the concentrations of A and B, m is the order of the reaction with respect to A, and n is the order of the reaction with respect to B.
02

Analyze the given data

Unfortunately, the data for this problem hasn't been provided. We need the initial concentrations of both A and B, as well as corresponding initial rates. In order to proceed, I will provide the following example data: | Experiment | Initial [A] (M) | Initial [B] (M) | Initial rate (M/s) | |------------|----------------|----------------|--------------------| | 1 | 0.1 | 0.1 | 0.25 | | 2 | 0.2 | 0.1 | 0.5 | | 3 | 0.1 | 0.2 | 1.0 | Now, we will use this data to calculate the reaction orders and rate constant.
03

Determine the order of the reaction with respect to A

To determine the order of the reaction with respect to A, we will compare Experiments 1 and 2, in which we doubled the concentration of A and kept the concentration of B constant. Observe how this affects the rate. From Experiment 1: \(rate_1 = k[A_1]^m[B_1]^n = k(0.1)^m(0.1)^n\) From Experiment 2: \(rate_2 = k[A_2]^m[B_2]^n = k(0.2)^m(0.1)^n\) Now, we will create a ratio of \(rate_2\) to \(rate_1\): \[\frac{rate_2}{rate_1} = \frac{(0.2)^m}{(0.1)^m}\] Using the provided initial rates: \(\frac{0.5}{0.25} = 2 = 2^m\) Thus, m = 1, so the reaction order with respect to A is 1.
04

Determine the order of the reaction with respect to B

To determine the order of the reaction with respect to B, we will compare Experiments 1 and 3, in which we doubled the concentration of B and kept the concentration of A constant. Observe how this affects the rate. From Experiment 1: \(rate_1 = k[A_1]^m[B_1]^n = k(0.1)^m(0.1)^n\) From Experiment 3: \(rate_3 = k[A_3]^m[B_3]^n = k(0.1)^m(0.2)^n\) Now, we will create a ratio of \(rate_3\) to \(rate_1\): \[\frac{rate_3}{rate_1} = \frac{(0.2)^n}{(0.1)^n}\] Using the provided initial rates: \(\frac{1.0}{0.25} = 4 = 2^n\) Thus, n = 2, so the reaction order with respect to B is 2.
05

Calculate the value of the rate constant k

Now that we have determined the reaction order with respect to A and B, we can calculate the rate constant k. From our previous findings, we know that the reaction is first-order with respect to A and second-order with respect to B. Thus, its rate law can be written as: \[rate = k[A]^1[B]^2\] We will use the data from Experiment 1 for our calculation: \[rate_1 = k[A_1]^1[B_1]^2\] \[0.25 = k(0.1)(0.1^2)\] \[0.25 = k(0.1)(0.01)\] Now, solve for k: \[k = \frac{0.25}{0.001} = 250\] So, the rate constant for the reaction is 250 M²/s.

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Most popular questions from this chapter

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