Problem 92 - Chemistry The Central Science In Si Units

Question

You perform a series of experiments for the reaction \(\mathrm{A} \rightarrow 2 \mathrm{~B}\) and find that the rate law has the form, rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) The rate increases by a factor of \(6.25,\) when \([A]_{0}\) is increased by a factor of \(2.5 .\) (b) There is no rate change when \([A]_{0}\) is increased by a factor of \(4 .(\mathbf{c})\) The rate decreases by a factor of \(1 / 2\), when \([A]\) is cut in half.

Step-by-Step Solution

Answer

(a) \(x = 2\), (b) \(x = 0\), (c) \(x = 1\).

Step 1: Introduction to the problem

We need to analyze the experimental conditions to determine the exponent \(x\) in the rate law, which is defined as \(\text{rate} = k[A]^x\). Given the changes in concentration and corresponding changes in rate, we will calculate \(x\) for each scenario.

Step 1: Calculate x for Case (a)

In case (a), the rate increases by a factor of 6.25 when \([A]_0\) increases by a factor of 2.5. Set up the equation: \[\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^x = 6.25\] \[\frac{[A]_2}{[A]_1} = 2.5\] \[2.5^x = 6.25\] Solving this equation: \(x = \log_{2.5}(6.25)\). Since \(6.25 = 2.5^2\), \(x = 2\).

Step 2: Calculate x for Case (b)

In case (b), there is no change in the rate when \([A]_0\) is increased by a factor of 4. This implies that the rate is independent of \([A]\), meaning \(x = 0\).

Step 3: Calculate x for Case (c)

In case (c), the rate decreases by a factor of \(\frac{1}{2}\) when \([A]\) is reduced by half. Set up the equation: \[\frac{\text{rate}_2}{\text{rate}_1} = \left( \frac{[A]_2}{[A]_1} \right)^x = \frac{1}{2}\] \[\frac{[A]_2}{[A]_1} = \frac{1}{2}\] Substitute and solve: \(\left(\frac{1}{2}\right)^x = \frac{1}{2}\) implies \(x = 1\).

Key Concepts

Reaction Kinetics

Reaction kinetics plays a crucial role in understanding how fast reactions occur. It deals with the rates of chemical reactions and the factors affecting these rates. When we talk about rate laws within reaction kinetics, we refer to equations describing how concentration influences the rate of reaction.
For a reaction like \(\mathrm{A} \rightarrow 2 \mathrm{~B}\), the rate law can be written in the form:

• \(\text{rate} = k[\mathrm{A}]^x\), where \(k\) is the rate constant and \(x\) is the reaction order with respect to \(\mathrm{A}\).

The value of \(x\) indicates how sensitive the rate is to changes in concentration.

• If \(x=0\), the rate doesn't depend on the concentration of \(\mathrm{A}\), indicating zero-order kinetics.
• If \(x=1\), the rate changes linearly with concentration changes, which means it's first-order.
• If \(x=2\), the rate is proportional to the square of the concentration, indicating second-order kinetics.

To determine \(x\), we perform experiments where we vary the concentration and measure how the rate changes. Understanding these concepts helps us predict how reactions can be controlled or optimized in various fields like chemical manufacturing or pharmaceuticals.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain unchanged over time. This occurs when the rates of the forward and reverse reactions are equal. In the context of the rate law \(\text{rate} = k[\mathrm{A}]^x\), we initially observe that the formation of \(\mathrm{B}\) from \(\mathrm{A}\) follows a rate dependent on the concentration of \(\mathrm{A}\).
At equilibrium, the forward reaction rate matches the reverse rate, resulting in no net change in concentrations. This balance explains why sometimes increasing concentrations stops affecting rates, as seen in case (b) where \(x=0\) signifies that beyond a certain concentration, the reaction rate does not increase further.

• Equilibrium doesn’t mean reactions stop; they continue to occur at equal rates.
• Understanding equilibrium helps to understand how reactions shift in response to concentration changes, temperature, or pressure changes.

In practice, understanding chemical equilibrium is essential in industries where reaction yields need maximization or minimization. It helps chemists control reactions to avoid unwanted side-products or increase efficiency.

Concentration Effect

The concentration effect refers to the impact that changes in concentration have on the reaction rate. In the context of the given rate law, \(\text{rate} = k[\mathrm{A}]^x\), the exponent \(x\) exemplifies this effect perfectly. The relation between concentration changes and rate changes is a fundamental concept in kinetics.

• Case (a) shows that if the concentration of \(\mathrm{A}\) is increased by a factor of 2.5, the rate increases by a factor of 6.25, indicating a second-order reaction.
• In case (b), increasing \([A]_0\) by a factor of 4 with no change in rate suggests zero-order kinetics, where the rate is independent of concentration.
• In case (c), halving the concentration and the rate also halves shows a first-order dependence on concentration.

By examining these examples, we see that understanding concentration effects allows prediction of how altering concentrations affect reaction rates. This understanding is crucial for optimizing reaction conditions in experimental and industrial setups. It ensures that reactions proceed at desired rates, which is vital for cost-effectiveness and resource efficiency.