Question

The following data pertains to the reaction between A and B $$ \begin{array}{llll} \hline \text { S. } & {[\mathrm{A}]} & {[\mathrm{B}]} & \text { Rate } \\ \text { No. } & \mathrm{mol} \mathrm{L}^{-1} & \mathrm{~mol} \mathrm{~L}^{-1} & \mathrm{Mol} \mathrm{L}^{-1} \mathrm{t}^{-1} \\ \hline 1 & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\ 2 & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-4} \\ \hline \end{array} $$ Which of the following inferences are drawn from the above data? (1) Rate constant of the reaction is \(10^{-4}\) (2) Rate law of the reaction is k [A][B] (3) Rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: a. 1 and 3 b. 2 and 3 c. 1 and 2 d. 1,2 and 3

Short answer

The correct answer is d. 1, 2, and 3.

Step-by-step solution

Identify the Rate Law Expression

The rate law expression for a reaction describes how the rate depends on the concentration of the reactants. For a reaction between species A and B, the rate law can be expressed as \( \text{Rate} = k [A]^m [B]^n \), where \( m \) and \( n \) are the orders of the reaction with respect to A and B, respectively, and \( k \) is the rate constant.

Determine the Reaction Orders

Comparing experiments 1 and 2, [B] is constant and [A] doubles from \(1 \times 10^{-2}\) to \(2 \times 10^{-2}\). The rate also doubles from \(2 \times 10^{-4}\) to \(4 \times 10^{-4}\). This suggests the reaction is first-order with respect to A. Next, compare experiments 2 and 3, where [A] is constant and [B] doubles from \(2 \times 10^{-2}\) to \(4 \times 10^{-2}\). The rate again doubles from \(4 \times 10^{-4}\) to \(8 \times 10^{-4}\), indicating the reaction is first-order with respect to B.

Establish the Rate Law

Since both reactants influence the rate proportionally when doubled, the rate law expression based on the observations is \( \text{Rate} = k [A][B] \), which confirms inference (2).

Calculate the Rate Constant

Using the rate law \( \text{Rate} = k [A][B] \) and substituting the values from experiment 1: \[ 2 \times 10^{-4} = k (1 \times 10^{-2})(2 \times 10^{-2}) \]. Solving for \( k \), we get: \[ k = \frac{2 \times 10^{-4}}{2 \times 10^{-4}} = 10^{-4} \]. Hence, inference (1) is also correct.

Verify Inference About Rate Increase

Re-examine the changes in rate when concentrations of A and B are doubled simultaneously. When [A] goes from 1 to 2 and [B] goes from 2 to 4, comparing experiment 1 and 3, the rate increases from \(2 \times 10^{-4}\) to \(8 \times 10^{-4}\), which is exactly four times. This confirms inference (3).

Key concepts

Understanding Rate Law

In the study of reaction kinetics, a "Rate Law" plays a crucial role. It describes how the speed, or rate, of a chemical reaction depends on the concentrations of the reactants involved. For a given reaction involving substances A and B, the rate law can be generally written as \( \text{Rate} = k [A]^m [B]^n \). Here, \( k \) is the rate constant, while \( m \) and \( n \) represent the reaction orders for A and B, respectively.
One of the main uses of the rate law is to predict how varying the concentration of one or more substances can affect the rate of reaction. This can be extremely helpful in optimizing conditions in chemical manufacturing or experimental settings.
When determining the rate law for a reaction, experimental data is essential. By analyzing how different concentrations affect reaction rates, one can infer the specific rate law applicable to that reaction.

Discovering Reaction Order

"Reaction Order" reveals how the rate of a chemical reaction is influenced by the concentration of its reactants. Reaction orders can be integers, fractions, or zero. Each reactant in a reaction may have its own order, and the overall reaction order is the sum of these individual orders. For example, in the rate law expression \( \text{Rate} = k [A]^m [B]^n \), the reaction order with respect to A is \( m \), and with respect to B is \( n \).
Determine reaction orders through methodical experimentation. By observing how changes in concentration affect the overall rate, the reaction order for each reactant can be established. If doubling the concentration of a reactant causes the rate to double, the reaction is first-order with respect to that reactant. Likewise, if quadrupling the rate suggests a second-order dependence. Each observation adds clarity to the complete reaction mechanism.
Recognizing these orders assists in predicting how reaction conditions influence reaction speed, crucial for efficient process design in industries.

The Role of Rate Constant in Kinetics

The "Rate Constant" \( k \) is a fundamental factor in chemical kinetics, influencing the speed of reactions. It appears in the rate law and remains constant for a given reaction under constant temperature and pressure. Its units can vary, depending on the overall reaction order.
The value of \( k \) is determined experimentally, by solving the rate law equation with known concentrations and rates. For instance, if a reaction's rate law is \( \text{Rate} = k [A][B] \), then using experimental data, \( k \) can be found by rearranging and solving the equation: \( k = \frac{\text{Rate}}{[A][B]} \).
The rate constant also reflects the reaction's intrinsic properties, including temperature effects. According to the Arrhenius equation, \( k \) increases with temperature, suggesting that higher temperatures accelerate reactions. Therefore, \( k \) is crucial for interpreting how environmental factors affect reaction rates and aid in scaling processes for industrial applications.