Problem 55 - Chemistry and Chemical Reactivity

Question

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

Step-by-Step Solution

Answer

The reaction rate doubles.

Step 1: Identify Initial Rate Equation

We start with the given rate equation for the reaction: \( \text{Rate} = k[A]^2[B] \). Here, \( k \) is the rate constant, \([A]\) is the concentration of substance \( A \), and \([B]\) is the concentration of substance \( B \).

Step 2: Define Changes in Concentrations

According to the problem, the concentration of \([A]\) is doubled, so it becomes \(2[A]\). The concentration of \([B]\) is halved, so it becomes \(\frac{1}{2}[B]\).

Step 3: Substitute Changes in Rate Equation

Substitute the new concentrations into the rate equation: \( \text{New Rate} = k(2[A])^2(\frac{1}{2}[B]) \).

Step 4: Calculate New Rate Expression

Simplify the equation: \((2[A])^2 = 4[A]^2\) and \(\frac{1}{2}[B]\). Therefore, \( \text{New Rate} = k \cdot 4[A]^2 \cdot \frac{1}{2}[B] = 2k[A]^2[B] \).

Step 5: Determine Change in Reaction Rate

Compare \(2k[A]^2[B]\) to the original \(k[A]^2[B]\): The new rate is twice the original rate. Therefore, the reaction rate doubles when \([A]\) is doubled and \([B]\) is halved.

Key Concepts

Rate Law

Rate laws are key to understanding how different factors affect the speed of chemical reactions. They express the reaction rate in terms of the concentration of reactants and a rate constant, represented in an equation. In our example, the rate equation is given by \( \text{Rate} = k[A]^2[B] \). Here, \([A]\) and \([B]\) are the concentrations of reactants \(A\) and \(B\), respectively, and \(k\) is the rate constant.
Understanding the format of a rate law is crucial because it tells us the order of the reaction with respect to each reactant. In this case, the reaction is second order with respect to \(A\) (since its exponent is 2) and first order with respect to \(B\), making it a third-order reaction overall. Changes in these concentrations dynamically influence the reaction rate, which is why comprehending each component of this equation is essential.
Sometimes, the rate law must be determined experimentally because it cannot be deduced from the balanced chemical equation. Thus, it is vital to understand how to use the rate law to predict how changes in concentration affect reaction rates.

Concentration Changes

When we talk about concentration changes, we're focusing on how varying the amount of reactant in a given volume affects the reaction rate. In our exercise, if the concentration of \(A\) is doubled, it changes from \([A]\) to \(2[A]\). Similarly, if the concentration of \(B\) is halved, it changes from \([B]\) to \(\frac{1}{2}[B]\).
Such changes directly impact the rate of reaction since the rate law depends on concentration. Using our original expression \( \text{Rate} = k[A]^2[B] \), substituting \([A] = 2[A]\) and \([B] = \frac{1}{2}[B]\), we get:

• \((2[A])^2 = 4[A]^2\)
• \(\frac{1}{2}[B]\)

Integrating these into the rate equation leads to \( \text{New Rate} = 4k[A]^2 \cdot \frac{1}{2}[B] = 2k[A]^2[B]\), indicating that the reaction rate doubles.
Thus, understanding how concentration changes affect the rate is pivotal in controlling and optimizing reaction speeds in various chemical processes.

Rate Constant

The rate constant, \( k \), is a crucial component of the rate law, serving as a proportionality factor in the rate equation. It is unique to each reaction and is affected by factors such as temperature and the presence of a catalyst, but remains constant with changes in concentration.
In our reaction rate expression \( \text{Rate} = k[A]^2[B] \), \( k \) dictates the speed of the reaction under a specific set of conditions. It encapsulates the inherent characteristics of the reaction, such as its reaction mechanism and the activation energy required for the reaction to proceed.
The value of \( k \) helps determine how fast or slow a reaction generally goes under given conditions, making it a valuable parameter for comparison across different reactions. Furthermore, it remains unchanged in the process of varying reactant concentrations in an isolated system, like we demonstrated in our problem where doubling \([A]\) and halving \([B]\) changed the rate but left \( k \) untouched.
This constancy of the rate constant in response to concentration changes underscores its role as a steadfast anchor in reaction kinetics.