Question

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2} .\) (a) If [A] is doubled, how will the rate change? Will the rate constant change? Explain. (b) What are the reaction orders for \(A\) and \(B\) ? What is the overall reaction order? (c) What are the units of the rate constant?

Short answer

Doubling the concentration of A does not change the rate or the rate constant. The reaction orders for A and B are 0 and 2, respectively, and the overall reaction order is 2. The units of the rate constant are \( \frac{1}{M \times s} \).

Step-by-step solution

Understand the rate law

The given rate law is Rate \(= k[\mathrm{B}]^2\). This tells us that the reaction rate depends on the concentration of B raised to the power of 2, and a constant k. Note that there is no dependence on the concentration of A.

Evaluate the effect of doubling [A]

Since the rate law does not depend on the concentration of A, doubling the concentration of A will have no effect on the rate of the reaction. The rate will remain the same.

Determine if the rate constant changes

As the name suggests, the rate constant, k, remains constant for a given reaction under specific conditions. Therefore, doubling the concentration of A will not affect the rate constant. #a: Answer# Doubling the concentration of A does not change the rate or the rate constant. #b: Determine the reaction orders for A and B, and the overall reaction order.#

Identify the reaction order for A

From the given rate law, Rate \(= k[\mathrm{B}]^2\), we see that there is no mention of concentration of A. This indicates that the reaction order for A is 0.

Identify the reaction order for B

The rate law given is Rate \(= k[\mathrm{B}]^2\). The exponent in the rate law represents the reaction order with respect to B. Here, the exponent is 2. So, the reaction order for B is 2.

Calculate the overall reaction order

The overall reaction order is the sum of the reaction orders of all the reactants. In this case, the reaction orders for A and B are 0 and 2, respectively. Overall reaction order = 0 + 2 = 2. #b: Answer# The reaction orders for A and B are 0 and 2, respectively, and the overall reaction order is 2. #c: Determine the units of the rate constant.#

Write the rate law with respects to units

We need to consider the rate law Rate \(= k[\mathrm{B}]^2\) and the units associated with each variable. The units of rate are molarity/time (M/s, M/min, etc.), and the units of concentration are molarity (M).

Determine the units of the rate constant, k

Plug the units for rate and concentration into the rate law and solve for k: Rate (M/s) = k ([M])^2 k = (M/s) / (M^2) The units of the rate constant k can be simplified as follows: k = \( \frac{M/s}{M^2} \) = \( \frac{1}{M \times s} \) #c: Answer# The units of the rate constant are \( \frac{1}{M \times s} \).

Key concepts

Understanding Chemical Kinetics and Rate Law

Chemical kinetics is the study of reaction rates, how they change under various conditions, and the steps or mechanisms through which they occur. A fundamental concept in chemical kinetics is the rate law, which provides a mathematical relationship between the rate of a reaction and the concentration of its reactants.

The rate law for a chemical reaction can be written in a general form as:\[\begin{equation}Rate = k[A]^m[B]^n\text{(. . . )}\end{equation}\]In this equation, \(k\) is the rate constant, \(A\) and \(B\) are the reactants, and \(m\) and \(n\) represent the reaction orders with respect to each reactant, respectively. The rate law tells us not only which reactants affect the rate but also how sensitive the rate is to changes in those reactants' concentrations.

For example, imagine a scenario where the reaction rate does not change when the concentration of reactant \(A\) is altered. This would suggest that the reaction is zero-order with respect to \(A\), indicating that changes in \(A\)'s concentration have no effect on the reaction rate. This concept is crucial for predicting how differing reactant concentrations can influence the overall speed at which a reaction proceeds.

Deciphering Reaction Order

A reaction order is essentially an exponent in the rate law equation, reflecting the power to which the concentration of a reactant is raised. For instance, in the rate law\[\begin{equation}Rate = k[B]^2\text{(. . . )}\end{equation}\]the reaction is second-order concerning \(B\), as seen by the exponent of 2. This implies that if the concentration of \(B\) is doubled, the rate of the reaction would increase by a factor of four, assuming all other factors are kept constant.

In cases where the concentration of a reactant does not appear in the rate law, like reactant \(A\) in our example, we say the reaction is zero-order with respect to that reactant. The overall reaction order is the sum of the orders with respect to each reactant involved in the rate-determining step. It reveals the overall dependence of the rate on the concentration of all reactants.

Interpreting the Rate Constant

The rate constant \(k\) is more than just a number; it is a reflection of the intrinsic reaction rate and remains unaffected by changes in reactant concentrations. It is specific to a particular reaction at a given temperature; different reactions or temperatures will feature different rate constants.

The units of the rate constant provide insightful details about the reaction mechanism and order. For a second-order rate law like\[\begin{equation}Rate = k[B]^2\text{(. . . )}\end{equation}\]the units for \(k\) are \( M^{-1} s^{-1} \), derived from dividing the units of rate (such as \( M s^{-1} \) or \(mol/L \times \frac{1}{s}\)) by the square of the concentration units (\[M^2\]). These units are essential for calculating and predicting the rates of chemical reactions and for studying the impact of variables such as temperature or catalyst presence on the reaction rate.