Problem 8 - Chemistry and Chemical Reactivity

Question

A reaction has the experimental rate equation Rate \(=k[\mathrm{A}]^{2} .\) How will the rate change if the concentration of \(\mathrm{A}\) is tripled? If the concentration of \(A\) is halved?

Step-by-Step Solution

Answer

Tripling \([\mathrm{A}]\) increases the rate by 9; halving \([\mathrm{A}]\) decreases the rate by 4.

Step 1: Understanding the Rate Law

The given rate law for the reaction is \( \text{Rate} = k[\mathrm{A}]^2 \). This means the rate of the reaction depends on the square of the concentration of \( \mathrm{A} \).

Step 2: Calculating Rate Change If A is Tripled

If the concentration of \( \mathrm{A} \) is tripled, \([\mathrm{A}]\) becomes \(3\times[\mathrm{A}]\). Substitute into the rate equation:\[ \text{Rate} = k(3[\mathrm{A}])^2 = k \cdot 9[\mathrm{A}]^2. \] Thus, the rate increases by a factor of 9.

Step 3: Calculating Rate Change If A is Halved

If the concentration of \( \mathrm{A} \) is halved, \([\mathrm{A}]\) becomes \(\frac{1}{2}\times[\mathrm{A}]\). Substitute into the rate equation:\[ \text{Rate} = k\left(\frac{1}{2}[\mathrm{A}]\right)^2 = k \cdot \frac{1}{4}[\mathrm{A}]^2. \] Thus, the rate is reduced by a factor of 4.

Key Concepts

Reaction Kinetics

Reaction kinetics is a fascinating study, diving into how chemical reactions occur and how fast they progress. The speed of a reaction is pivotal in both natural and industrial processes. Reaction kinetics helps us to understand how different factors influence the rate at which reactants are converted into products. In our exercise, the rate is influenced by the concentration of substance A. This dependency showcases the fundamental idea that the speed of a chemical reaction is not static. Instead, it adjusts in response to changes in the reaction conditions. By analyzing the rate law, which depicts the relationship between concentration and reaction rate, we can predict and control reactions effectively.

Rate Equation

A rate equation is a mathematical expression that links the rate of a reaction to the concentration of its reactants. It allows chemists to quantify how changes in concentration influence the speed of reactions. In our scenario, the rate equation is given by \( ext{Rate} = k[ ext{A}]^2 \). Here, \( k \) is the rate constant, a unique value for each reaction at a given temperature.

• \([\text{A}]\) represents the concentration of reactant A.
• The exponent, 2, indicates the order of the reaction concerning A, suggesting that the rate is highly sensitive to changes in A's concentration.

When solving problems using rate equations, it’s crucial to correctly interpret these components to understand how rate changes will occur with varying reactant concentrations.

Concentration Effect

Concentration plays a vital role in determining how quickly a reaction proceeds. The rate equation in our example shows this effect clearly, as any change in \([\mathrm{A}]\) significantly alters the reaction rate. Consider the effect when \([\mathrm{A}]\) is changed:

• If \([\text{A}]\) is tripled, the new concentration becomes \(3[\text{A}]\). Plugging this into the rate equation results in the rate increasing by a factor of 9, because \([\text{A}]^2\) now becomes \(9[\text{A}]^2\).


• Conversely, if \([\text{A}]\) is halved, the concentration becomes \(\frac{1}{2}[\text{A}]\). The rate then decreases by a factor of 4, as \([\text{A}]^2\) changes to \(\frac{1}{4}[\text{A}]^2\).

This demonstrates the power of concentration changes on reaction dynamics, allowing chemists to manipulate reaction speeds through careful adjustment of reactant concentrations.