Question

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 \(R_{1} ?\)

Short answer

The reaction is first order with respect to A, second order with respect to B, and negative first order with respect to C. The value of \(R_{5}\) in terms of \(R_{1}\) is one half, (\(R_{5} = \frac{1}{2} * R_{1}\)).

Step-by-step solution

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.

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.

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.

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.

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, \(R_5 = \frac{1}{2} * R_1\).

Key concepts

Rate of Reaction

In chemistry, the rate of reaction is a fascinating topic that describes how rapidly or slowly a reaction occurs. It measures the speed at which reactants are converted into products. This rate can tell us a lot about how different conditions, such as temperature or concentration, impact reactions. Think of it like cooking; some ingredients might mix faster than others based on how finely they are chopped or how warm the oven is!

In the given reaction, the rate is influenced by changes in the concentrations of the reactants A, B, and C. When reacting, these chemicals can speed up or slow down the process, and we capture this as the initial rate. It's vital to remember that a reaction's rate itself can offer insights into the reaction mechanism—the step-by-step sequence of elementary reactions by which overall chemical change occurs.

By carefully observing changes in initial rates (like in the table given in the exercise), we can deduce the influence each reactant has on the reaction. For instance, reducing A's concentration by half results in halving the rate suggests A is first-order. This explanation helps students grasp why experiment data is crucial in understanding the underlying principles of chemistry.

Reaction Kinetics

Reaction kinetics digs deeper into understanding the detailed processes behind reaction rates. It's like figuring out a car's speed by considering not just the fuel, but also the engine design and terrain. In the study of reaction kinetics, we focus on how the concentration of reactants and other factors influence the rate.

In the exercise, we observe certain trends: halving the concentration of a reactant and seeing how it affects the rate gives clues about the reaction order. These orders are like the rules dictating how reactants relate to the rate.

Understanding kinetics involves terms such as:

• **First-order reactions:** These depend linearly on one reactant's concentration (e.g., A in the exercise).
• **Second-order reactions:** The rate is proportional to the square of a concentration, or the product of two concentrations (such as B in the exercise).
• **Negative order reactions:** They indicate unusual behavior like a rate increasing when a concentration decreases, which happens with C in the scenario.

Mastering reaction kinetics equips us to predict reaction behavior, optimize conditions, and even design experiments to probe unknown reactions.

Chemical Equations

Chemical equations are the backbone of chemistry, providing a shorthand way to describe chemical reactions. They're like a recipe listing ingredients and the dish they make, but with reactants and products.

In the exercise's chemical equation, we have: \( A + 2B + C \rightarrow 2D + E \)This equation shows what's consumed and produced, and the stoichiometry of each component—the numbers in front indicating how many molecules are involved.

Understanding a chemical equation involves:

• **Reactants:** Substances initially present, like A, B, and C here.
• **Products:** Substances formed, shown on the right side (2D and E).
• **Balancing equations:** Making sure the number of each type of atom is the same on both sides.

A balanced chemical equation is crucial, as it allows chemists to calculate precisely how much of each reactant is needed and how much product is expected, guiding lab work and commercial production.