Question

For the following first-order competing reaction: $$ \begin{array}{l} \text { A + Reagent } \rightarrow \text { Product } \\ \text { B + Reagent } \rightarrow \text { Product } \end{array} $$ the ratio of \(K_{1} / K_{2}\), if only \(50 \%\) of ' \(\mathrm{B}\) ' will have been reacted when \(94 \%\) of ' \(\mathrm{A}\) ' has been reacted is \((\log 2=0.3, \log 3=0.48)\) (a) \(4.06\) (b) \(0.246\) (c) \(8.33\) (d) \(0.12\)

Short answer

The closest calculated ratio of the rate constants \(K_{1} / K_{2}\) is (a) 4.06.

Step-by-step solution

Understand the first-order reaction kinetics

First-order reaction kinetics state that the rate of reaction is directly proportional to the concentration of the reactant. The general form of a first-order rate equation is: \( rate = k \times [\text{Reactant}] \), where \( k \) is the rate constant.

Apply first-order reaction mathematics to Reactant A

Given that 94% of A has reacted, this implies that 6% remains unreacted. For a first-order reaction, the time taken to react depends on the initial concentration and the rate constant. The formula to relate these is \( t = \frac{1}{k_1} \times \text{ln} \frac{\text{Initial Concentration of A}}{\text{Remaining Concentration of A}} \).

Apply first-order reaction mathematics to Reactant B

Similarly, given that 50% of B has reacted, 50% remains unreacted. The relationship between time, rate constant for B, and concentrations is \( t = \frac{1}{k_2} \times \text{ln} \frac{\text{Initial Concentration of B}}{\text{Remaining Concentration of B}} \).

Equate the times for both reactions to find the ratio \( K_{1} / K_{2} \)

Since the percentage of reactants reacted refers to the same time frame for both A and B, we can equate the two expressions for time: \( \frac{1}{k_1} \times \text{ln} \frac{1}{0.06} = \frac{1}{k_2} \times \text{ln} \frac{1}{0.50} \).

Insert the logarithm values

Using the provided logarithm approximations, we can replace the natural logarithms with their values. Due to the property of logarithms that \( \text{ln}(x^{-1}) = -\text{ln}(x) \), the equation becomes: \( \frac{1}{k_1} \times -(0.3) = \frac{1}{k_2} \times \text{ln}2 \), and using the value of \( \text{ln}2 = 0.3 \), we get: \( \frac{1}{k_1} \times -(0.3) = \frac{1}{k_2} \times (0.3) \).

Calculate the ratio \( K_{1} / K_{2} \)

Solving the equation for the ratio \( K_{1} / K_{2} \) gives us: \( \frac{k_1}{k_2} = -1 \ \frac{k_1}{k_2} = \frac{0.3}{0.3} \), thus \( \frac{k_1}{k_2} = 1 \). This result needs to be revisited since it implies the two rate constants are equal, which does not match any of the provided answers. We must have made an error in previous steps.

Reevaluate the logarithms and calculations

The proper use of logarithms and the provided logarithmic values implies: \( \frac{1}{k_1} \times \text{ln}(1/0.06) = \frac{1}{k_2} \times \text{ln}(1/0.50) \). Substituting in the given logarithm values, we get: \( \frac{1}{k_1} \times \text{ln}(16.67) = \frac{1}{k_2} \times \text{ln}(2) \), which implies \( \frac{1}{k_1} \times (0.3 \times 4) = \frac{1}{k_2} \times 0.3 \). This simplifies to \( \frac{0.3 \times 4}{k_1} = \frac{0.3}{k_2} \) and the ratio \( \frac{k_1}{k_2} = 4 \).

Compare to the provided answers

Finally, we compare the calculated ratio to the provided multiple-choice answers. None of them match the value of 4 exactly, but the closest answer is (a) 4.06. Given that logical approximations were used for the logarithm values, answer (a) 4.06 is the closest to our calculated ratio.

Key concepts

Rate Constant

The rate constant is a crucial concept in reaction kinetics, serving as a bridge between the speed of a reaction and the concentrations of reactants. Mathematically, we express the rate of a first-order reaction as \( rate = k \times [\text{Reactant}] \), with \( k \) representing the rate constant.

The rate constant is distinctive for each reaction and under fixed conditions, encapsulating factors like temperature and the presence of a catalyst. In the case of first-order reactions, the rate constant helps determine how swiftly the concentration of the reactant diminishes over time, which is proportional to the current concentration of that reactant. A larger rate constant signifies a faster reaction under comparable conditions. The dimensionality of the rate constant varies according to the overall reaction order, and for a first-order reaction, it is typically expressed in \( s^{-1} \).

Reaction Rate

Reaction rate can be visualized as the speedometer of a chemical reaction, quantifying how quickly reactants convert into products. For first-order reactions, the rate depends on the concentration of one reactant. Simply put, as you have more reactant molecules, there are more frequent collisions and opportunities for the reaction to occur, driving the rate up.

The first-order rate equation \( rate = k \times [\text{Reactant}] \) seamlessly portrays this relationship. Understanding that for each reduction in reactant concentration by half, the reaction rate also halves, allows one to anticipate the reaction's progression. This dependency paves the way for solving various quantitative problems in chemistry, such as predicting how long it will take for a certain percentage of the reactant to transform into the product.

Competing Reactions

Competing reactions often occur in chemical scenarios where two or more reaction pathways are available for the reactants. In our exercise, both \( A \) and \( B \) react with a common reagent to produce the same product, but at different rates governed by their respective rate constants \( k_1 \) and \( k_2 \).

Understanding competing reactions is crucial for controlling product distribution in chemical manufacturing. The ratio of rate constants, or the selectivity, is a measure expressing which reaction pathway is favored. In the given problem, by analyzing the time it takes for certain percentages of \( A \) and \( B \) to react, we can calculate the ratio of \( K_{1} / K_{2} \), defining the selectivity between the two pathways.

When competing reactions are of the same order, we can equate the time necessary for a specific conversion of reactants since they share the same reaction mechanism and simply compare the two rates. In more complex scenarios, controlling the conditions to favor one reaction over another becomes an essential skill in chemistry.