Rate Laws from Experimental Data: Initial Rates Method. The reaction $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$\text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t}$$ $$ \begin{array}{ccc} {[\mathrm{NO}]_{0}} & {\left[\mathrm{Cl}_{2}\right]_{0}} & \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ 0.10 & 0.10 & 0.18 \\ 0.10 & 0.20 & 0.36 \\ 0.20 & 0.20 & 1.45 \end{array} $$ a. What is the rate law? b. What is the value of the rate constant?

Step by step solution

01

Determine the order of the reaction with respect to NO

Compare experiment 1 and 3, where the initial concentration of Cl2 is constant. This will help us determine the effect of changing the concentration of NO on the rate of the reaction. Experiment 1: [NO] = 0.10 M, [Cl2] = 0.10 M, Initial Rate = 0.18 M/min Experiment 3: [NO] = 0.20 M, [Cl2] = 0.20 M, Initial Rate = 1.45 M/min The concentration of NO doubles while the concentration of Cl2 and the rate of the reaction are kept constant. Observe how much the initial rate increases when [NO] doubles: $$\frac{1.45}{0.18} = 8.06 \approx 8$$ The rate increases by a factor of 8 when the concentration of NO doubles, which indicates that the reaction is second order with respect to NO.

02

Determine the order of the reaction with respect to Cl2

Compare experiment 1 and 2, where the initial concentration of NO is constant. This will help us determine the effect of changing the concentration of Cl2 on the rate of the reaction. Experiment 1: [NO] = 0.10 M, [Cl2] = 0.10 M, Initial Rate = 0.18 M/min Experiment 2: [NO] = 0.10 M, [Cl2] = 0.20 M, Initial Rate = 0.36 M/min The concentration of Cl2 doubles while the concentration of NO and the rate of the reaction are kept constant. Observe how much the initial rate increases when [Cl2] doubles: $$\frac{0.36}{0.18} = 2$$ The rate doubles when the concentration of Cl2 doubles, which indicates that the reaction is first order with respect to Cl2.

03

Write the rate law

Since the reaction is second order with respect to NO and first order with respect to Cl2, the rate law can be written as: $$\text { Rate } = k[\mathrm{NO}]^2[\mathrm{Cl}_2]$$ where k is the rate constant.

04

Calculate the rate constant, k

Use the rate law from step 3 and the experimental data from any of the experiments. In this case, let's use the data from experiment 1. Initial Rate = 0.18 M/min, [NO] = 0.10 M, [Cl2] = 0.10 M Substitute the values into the rate law: $$0.18 = k(0.10)^2(0.10)$$ Solve for k: $$k = \frac{0.18}{(0.10)^2(0.10)} = 180$$ The rate constant k is 180 L^2/mol^2 min. The final answers are: a. Rate law: \(\text { Rate } = k[\mathrm{NO}]^2[\mathrm{Cl}_2]\) b. Rate constant: k = \(180 L^2/mol^2 min\)

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