The following data are obtained for the initial rates of reaction in the reaction \(A+2B+C \longrightarrow 2 D+E.\) $$\begin{array}{lllll} \hline & \text { Initial } & \text { Initial } & & \\ \text { Expt } & \text { [A], M } & \text { [B],M } & \text { [C], M } & \text { Initial Rate } \\ \hline 1 & 1.40 & 1.40 & 1.00 & R_{1} \\ 2 & 0.70 & 1.40 & 1.00 & R_{2}=\frac{1}{2} \times R_{1} \\ 3 & 0.70 & 0.70 & 1.00 & R_{3}=\frac{1}{4} \times R_{2} \\ 4 & 1.40 & 1.40 & 0.50 & R_{4}=16 \times R_{3} \\ 5 & 0.70 & 0.70 & 0.50 & R_{5}=? \\ \hline \end{array}$$ (a) What are the reaction orders with respect to A, B, and C? (b) What is the value of \(R_{5}\) in terms of \(R_{1} ?\)

Step by step solution

01

Consider Experiments 1 and 2

The initial concentration of A is halved between these experiments, from 1.40 M to 0.70 M, while the concentrations of B and C remained constant. As a result, the initial rate \(R_2\) becomes half of \(R_1\). From this, one can infer that the reaction is first order with respect to A, as decreasing the concentration of A by half causes a similar decrease in the rate.

02

Consider Experiments 2 and 3

The initial concentration of B is halved between these experiments, from 1.40 M to 0.70 M, while the concentrations of A and C remained constant. As a result, the rate \(R_3\) is one-fourth of \(R_2\). This suggests that the reaction is second order with respect to B, because halving B's concentration quadruples the reduction.

03

Consider Experiments 3 and 4

The initial concentration of C (a reactant) is halved between these two experiments, from 1.00 M to 0.50 M, while A and B remain constant. The rate for Experiment 4, \(R_4\), is 16 times \(R_3\). Therefore, the reaction is negative first order with respect to C, meaning that halving C's concentration increases the rate 16 times.

04

Combine Orders

The overall reaction order is the sum of the orders with respect to each reactant. Therefore, the reaction is overall second order since it's first order in A, second order in B, and negative first order in C.

05

Calculate the value of \(R_5\)

To find the value of \(R_5\), compare experiments 4 and 5 as both have the same initial concentrations for A and C but a halving of B’s concentration. Because the reaction is second order with respect to B, halving the initial concentration of B from experiment 4 to 5 should make the rate one fourth in the rate from experiment 4, therefore \(R_5 = \frac{1}{4} * R_4 = \frac{1}{4} * 16 * R_3 = 4 * R_3\). Using experiment 3, we know that \(R_3 = \frac{1}{4} * R_2\). Therefore, \(R_5 = 4 * \frac{1}{4} * R_2 = R_2\). We also know that \(R_2 = \frac{1}{2} * R_1\), therefore, \(R_5 = \frac{1}{2} * R_1\).

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