For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is : [A] (mol L–1) [B] (mol L–1) Initial Rate (mol L–1S–1) 0.05 0.05 0.045 0.10 0.05 0.090 0.20 0.10 0.72

Engineering

Chemistry

Methods of Determination of Order of Reaction

Question

For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is :

[A] (mol L^–1)	[B] (mol L^–1)	Initial Rate (mol L^–1S^–1)
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

Rate = k [A]² [B]

Rate = k [A]² [B]²

Rate = k [A] [B]

Rate = k [A] [B]²

r = K[A]^x [B]^y

0.045 = K(0.05)^x (0.05)^y . . . . (1)

0.090 = K(0.10)^x (0.05)^y . . . . (2)

0.72 = K(0.20)^x (0.10)^y . . . . (3)

Dividing (1) by (2) we get

$\frac{0 .045}{0 .090} = {(\frac{0 .05}{0 .10})}^{x} \Rightarrow x = 1$

Dividing (2) by (3)

$\frac{0 .090}{0 .720} = {(\frac{0 .10}{0 .20})}^{x} (\frac{0 .05}{0 .10}) \Rightarrow y = 2$

To determine the rate law for the reaction $2 A + B \to C$ , we use the method of initial rates. The general form of the rate law is $Rate = k [A]^{m} [B]^{n}$ , where $m$ and $n$ are the orders with respect to A and B, respectively, and $k$ is the rate constant.

Step 1: Determine the order with respect to A (find m)

Compare experiments where [B] is constant. Look at Experiment 1 and 2.

Experiment 1: [A] = 0.05 M, [B] = 0.05 M, Rate = 0.045 M/s

Experiment 2: [A] = 0.10 M, [B] = 0.05 M, Rate = 0.090 M/s

Notice that [B] is the same in both, but [A] doubles (from 0.05 to 0.10). The rate also doubles (from 0.045 to 0.090).

Therefore, $\frac{Rate2}{Rate1} = \frac{0.090}{0.045} = 2$ and $\frac{[}{A} [A]_{1} = \frac{0.10}{0.05} = 2$ .

So, $2 = 2^{m}$ . This means $m = 1$ . The reaction is first order with respect to A.

Step 2: Determine the order with respect to B (find n)

Now we need to compare experiments where [A] is constant. No two experiments have the same [A], so we must use the general rate law ratio. Let's compare Experiment 2 and 3.

Experiment 2: [A] = 0.10 M, [B] = 0.05 M, Rate = 0.090 M/s

Experiment 3: [A] = 0.20 M, [B] = 0.10 M, Rate = 0.72 M/s

Write the ratio of the rate laws for these two experiments:

$\frac{Rate3}{Rate2} = \frac{k}{[} k [A]_{2}]^{m} [B]_{2}]^{n} = {(\frac{[}{A} [A]_{2})}^{m} {(\frac{[}{B} [B]_{2})}^{n}$

Substitute the known values. We found $m = 1$ .

$\frac{0.72}{0.090} = 8$

$\frac{[}{A} [A]_{2} = \frac{0.20}{0.10} = 2$

$\frac{[}{B} [B]_{2} = \frac{0.10}{0.05} = 2$

Now plug these into the equation:

$8 = 2^{1} 2^{n}$

$8 = 2 2^{n}$

$\frac{8}{2} = 4 = 2^{n}$

Since $2^{2} = 4$ , this means $n = 2$ . The reaction is second order with respect to B.

Step 3: Write the Rate Law

The order with respect to A is 1 and the order with respect to B is 2. Therefore, the rate law is:

$Rate = k [A]^{1} [B]^{2}$ or simply $Rate = k [A] [B]^{2}$ .

Final Answer: Rate = k [A] [B]²

Related Concepts and Formulae

1. Rate Law: An equation that relates the reaction rate to the concentrations of reactants. It is determined experimentally and cannot be deduced from the stoichiometry of the reaction alone. The general form for a reaction with reactants A and B is $Rate = k [A]^{m} [B]^{n}$ .

2. Method of Initial Rates: This is the most common method for experimentally determining the orders of reaction (m, n, etc.).

You measure the initial rate of reaction for several different starting concentrations of reactants.
You compare two experiments where the concentration of only one reactant changes while the others are held constant.
The ratio of the rates reveals the order with respect to the reactant that was changed.

3. Order of Reaction:

Zero Order: Rate is independent of the concentration of that reactant. Changing its concentration has no effect on the rate.
First Order: Rate is directly proportional to the concentration. Doubling the concentration doubles the rate.
Second Order: Rate is proportional to the square of the concentration. Doubling the concentration quadruples the rate.

4. Overall Order: The sum of the exponents (m + n) in the rate law. For our answer, Rate = k [A]¹[B]², the overall order is 1 + 2 = 3 (third order).

Initial Concentration (A)	Initial Concentration (B)	Initial rate of Formation of C (mol L^–S^–)
0.1 M	0.1 M	1.2 × 10^–3
0.1 M	0.2 M	1.2 × 10^–3
0.2 M	0.1 M	2.4 × 10^–3

Detailed Solution

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5