Problem 94 - Chemistry: The Central Science

Question

The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{X}]_{0}(M)} & {[\mathrm{Z}]_{0}(M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.25 & 0.25 & 4.0 \times 10^{1} \\ 2 & 0.50 & 0.50 & 3.2 \times 10^{2} \\ 3 & 0.50 & 0.75 & 7.2 \times 10^{2} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(X\) is \(0.75 M\) and that of \(Z\) is \(1.25 M ?\)

Step-by-Step Solution

Answer

(a) The rate law for this reaction is: Rate = \( k [\mathrm{X}]^2 [\mathrm{Z}] \) (b) The value of the rate constant is approximately \( 2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}} \) (c) The reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M is approximately \( 7.2 \times 10^3 \, \mathrm{M \, s^{-1}} \).

Step 1: Determine the order of the reaction with respect to reactant X

Compare Experiment 1 and Experiment 2, as both have a change in concentration for X but not for Z. The rate and concentration ratio can help us determine the order of the reaction with respect to X. For the change between Experiment 1 and Experiment 2: Change in concentration of X: \( \cfrac{[\mathrm{X}]_{0}\text{ (Experiment 2)}}{[\mathrm{X}]_{0}\text{ (Experiment 1)}} = \cfrac{0.50}{0.25} = 2 \) Change in rate: \( \cfrac{\mathrm{Rate} \cdot \mathrm{(Experiment 2)}}{\mathrm{Rate} \cdot \mathrm{(Experiment 1)}} = \cfrac{3.2 \times 10^2}{4.0 \times 10^1} = 8 \) Since the rate changes in 8 times while the concentration of X changes in 2 times, we can conclude that the reaction is second order with respect to X.

Step 2: Determine the order of the reaction with respect to reactant Z

Compare Experiment 2 and Experiment 3, as both have a change in concentration for Z but not for X. The rate and concentration ratio can help us determine the order of the reaction with respect to Z. For the change between Experiment 2 and Experiment 3: Change in concentration of Z: \( \cfrac{[\mathrm{Z}]_{0}\text{ (Experiment 3)}}{[\mathrm{Z}]_{0}\text{ (Experiment 2)}} = \cfrac{0.75}{0.50} = 1.5 \) Change in rate: \( \cfrac{\mathrm{Rate} \cdot \mathrm{(Experiment 3)}}{\mathrm{Rate} \cdot \mathrm{(Experiment 2)}} = \cfrac{7.2 \times 10^2}{3.2 \times 10^2} = 2.25 \) Since the rate changes in 2.25 times while the concentration of Z changes in 1.5 times, we can conclude that the reaction is first order with respect to Z.

Step 3: Determine the rate law and calculate the rate constant

From Steps 1 and 2, we know that the reaction is second order with respect to X and first order with respect to Z. Therefore, the rate law can be written as: Rate \( = k[\mathrm{X}]^2[\mathrm{Z}] \) Now let's use the data from Experiment 1 to calculate the rate constant k: \( 4.0 \times 10^1 = k (0.25)^2(0.25) \) \( k = \cfrac{4.0 \times 10^1}{(0.25)^2(0.25)} \) Calculate k: \( k \approx 2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}} \)

Step 4: Calculate the reaction rate for the given initial concentrations

Now that we have the rate law and the rate constant, we can calculate the reaction rate for the given initial concentrations of X = 0.75 M and Z = 1.25 M: Rate \( = k [\mathrm{X}]^2 [\mathrm{Z}] \) Rate \( = (2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}}) (0.75 \, \mathrm{M} )^2 (1.25 \, \mathrm{M}) \) Calculate the reaction rate: Rate \( \approx 7.2 \times 10^3 \, \mathrm{M \, s^{-1}} \) #Summary# (a) The rate law for this reaction is: Rate = \( k [\mathrm{X}]^2 [\mathrm{Z}] \) (b) The value of the rate constant is approximately \( 2.56 \times 10^3 \, \mathrm{M^{-3} s^{-1}} \) (c) The reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M is approximately \( 7.2 \times 10^3 \, \mathrm{M \, s^{-1}} \).