Question

The following data apply to the reaction $$ \mathrm{A}(g)+3 \mathrm{~B}(g)+2 \mathrm{C}(g) \longrightarrow \text { products } $$ $$ \begin{array}{cccc} \hline[\mathrm{A}] & {[\mathrm{B}]} & {[\mathrm{C}]} & \text { Rate } \\ \hline 0.20 & 0.40 & 0.10 & X \\ 0.40 & 0.40 & 0.20 & 8 X \\ 0.20 & 0.20 & 0.20 & X \\ 0.40 & 0.40 & 0.10 & 4 X \\ \hline \end{array} $$ Determine the rate law for the reaction.

Short answer

Question: Determine the rate law for the given reaction using the concentration-to-rate table provided. Answer: The rate law for the given reaction is: Rate = k * [A]^2 * [C]^5.

Step-by-step solution

Analyze the change in the rate and concentration for each reactant

In order to find the reaction orders of the reactants, we have to look at the given concentration-to-rate table and compare how the rate changes with the changes in the concentrations of A, B, and C.

Compare the concentrations and the rates of the reactants

Find a pair of rows in the table where the concentration of only one reactant changes and the other two remain constant. This will help us find the order of the reaction concerning that particular reactant.

Determine the order of A

To determine the order of A, compare rows 1 and 4, where the concentrations of B and C remain constant: Row 1: [A] = 0.20, [B] = 0.40, [C] = 0.10, Rate = X Row 4: [A] = 0.40, [B] = 0.40, [C] = 0.10, Rate = 4X When [A] doubles, the rate increases by a factor of 4. Therefore, the reaction is second-order with respect to A.

Determine the order of B

To determine the order of B, compare rows 1 and 3, where the concentrations of A and C remain constant: Row 1: [A] = 0.20, [B] = 0.40, [C] = 0.10, Rate = X Row 3: [A] = 0.20, [B] = 0.20, [C] = 0.20, Rate = X When [B] halves, the rate remains unchanged. Therefore, the reaction is zero-order with respect to B.

Determine the order of C

To determine the order of C, compare rows 1 and 2, where the concentrations of A and B remain constant: Row 1: [A] = 0.20, [B] = 0.40, [C] = 0.10, Rate = X Row 2: [A] = 0.40, [B] = 0.40, [C] = 0.20, Rate = 8X Since we already know the reaction is second-order with respect to A, let's find the rate when doubling [A] only: (0.20 / 0.40)^2 = 1/4 Therefore, the rate is equal to X/4 when doubling [A]. The rate in row 2 is 8X, so the change due to doubling [C] should be: (8X) / (X/4) = 32 When [C] doubles, the rate increases by a factor of 32. Therefore, the reaction is fifth-order with respect to C.

Write down the rate law

Now that we have determined the orders of the reactants, we can write down the rate law for the given reaction: Rate = k * [A]^2 * [B]^0 * [C]^5 Since [B]^0 equals 1, the final rate law is: Rate = k * [A]^2 * [C]^5.

Key concepts

Reaction Order

Understanding the reaction order is crucial when studying the dynamics of a chemical reaction. The reaction order indicates the power to which the concentration of a reactant is raised in the rate law. It reflects how the rate of the reaction is affected by the concentration of that reactant.

For instance, if a reaction is first-order with respect to a reactant, it means that the rate is directly proportional to the first power of that reactant's concentration. If the concentration doubles, the reaction rate will also double. On the other hand, in a zero-order reaction, the rate does not depend on the concentration of the reactant – it stays constant regardless of changes in concentration.

Rate of Reaction

The rate of reaction is the speed at which reactants are converted into products. It's measured by the change in concentration of reactants (or products) per unit time. Factors that can affect this rate include the concentration of reactants, their physical state, temperature, and the presence of catalysts.

To determine how a rate changes in response to the changes in reactant concentration, rate laws are employed. As in our exercise, by analyzing how the rate responds to concentration changes, we can deduce the reaction order and shape the rate law accordingly.

Chemical Kinetics

The field of chemical kinetics deals with studying the rates of chemical reactions and understanding the various factors that influence these rates. Kinetics can tell us a lot about a reaction's mechanism, that is, the steps that lead from reactants to products. By delving into kinetics, we uncover how different concentrations, temperature shifts, and catalysts affect the speed of a reaction, allowing us to control and optimize these reactions for industrial, pharmaceutical, or other practical applications.

Rate Law

A rate law, also known as a rate equation, connects the rate of a chemical reaction to the concentration of its reactants. It's an algebraic equation that conveys the relationship derived from experimental data, such as what we deduced in the exercise provided. The general form of a rate law is Rate = k * [A]ˆx * [B]ˆy * [C]ˆz, where [A], [B], and [C] represent the molar concentrations of reactants, 'x', 'y', and 'z' are the reaction orders, and 'k' is the rate constant, a proportionality factor unique to each reaction at a given temperature.

Reactant Concentration

The term reactant concentration refers to the amount of a given reactant present in a unit volume of the reaction mixture. It's usually expressed in moles per liter (Molarity, M). Reactant concentration plays a pivotal role in determining the reaction rate, as seen in our exercise. By altering concentrations, you can significantly change the rate of the reaction, more so if the reaction order with respect to that reactant is higher than one.

In practice, for any rate law determination, scientists often change the concentration of one reactant at a time while holding the others constant, as seen in steps 2 to 5 of the provided solution. This way, they can isolate and identify the effect of each reactant on the overall rate of reaction.