Questions: A reaction is performed to study the reaction of ICl with hydrogen: 2 ICl + H2 -> I2 + 2 HCl The following reaction rate data was obtained in four separate experiments. Experiment [ICl]0, M [H2]0, M Initial Rate, M s^-1 1 0.257 3.20 x 10^-2 4.87 x 10^-4 2 0.514 3.20 x 10^-2 9.74 x 10^-4 3 0.257 6.40 x 10^-2 1.95 x 10^-3 4 0.514 6.40 x 10^-2 3.89 x 10^-3 What is the rate law for the reaction and what is the numerical value of k ? Complete the rate law in the box below. Remember that an exponent of ' 1 ' is not shown and concentrations taken to the zero power do not appear. (Use k for the rate constant.) Rate = k= M^-2 s^-1

Solution Steps

To find the order of the reaction with respect to \([\mathrm{ICl}]\), compare experiments 1 and 2, where \([\mathrm{H}_2]\) is constant:

\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\mathrm{ICl}]_2}{[\mathrm{ICl}]_1}\right)^m \]

\[ \frac{9.74 \times 10^{-4}}{4.87 \times 10^{-4}} = \left(\frac{0.514}{0.257}\right)^m \]

\[ 2 = 2^m \]

Thus, \(m = 1\).

To find the order of the reaction with respect to \([\mathrm{H}_2]\), compare experiments 1 and 3, where \([\mathrm{ICl}]\) is constant:

\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[\mathrm{H}_2]_3}{[\mathrm{H}_2]_1}\right)^n \]

\[ \frac{1.95 \times 10^{-3}}{4.87 \times 10^{-4}} = \left(\frac{6.40 \times 10^{-2}}{3.20 \times 10^{-2}}\right)^n \]

\[ 4 = 2^n \]

Thus, \(n = 2\).

The rate law is:

\[ \text{Rate} = k [\mathrm{ICl}]^1 [\mathrm{H}_2]^2 \]

Using experiment 1 data to calculate \(k\):

\[ 4.87 \times 10^{-4} = k (0.257)^1 (3.20 \times 10^{-2})^2 \]

\[ 4.87 \times 10^{-4} = k \times 0.257 \times 1.024 \times 10^{-3} \]

\[ k = \frac{4.87 \times 10^{-4}}{0.257 \times 1.024 \times 10^{-3}} \]

\[ k = \frac{4.87 \times 10^{-4}}{2.63168 \times 10^{-4}} \]

\[ k \approx 1.850 \]

Final Answer