Question

In a first-order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the ratio of \(\mathrm{a} /(\mathrm{a}-\mathrm{x})\) was found to be 8 after 60 minutes. If the concentration is \(0.1 \mathrm{M}\) then the rate of reaction in moles of A reacted per minutes is (a) \(2.226 \times 10^{-5} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (b) \(3.466 \times 10^{-3}\) mol litre \(^{-1} \mathrm{~min}^{-1}\) (c) \(4.455 \times 10^{-3}\) mol litre \(^{-4} \mathrm{~min}^{-1}\) (d) \(5.532 \times 10^{-3}\) mol litre- \(^{-1} \min ^{-1}\)

Short answer

The rate of reaction is (b) \(3.466 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}\).

Step-by-step solution

Understand the Reaction Order

The problem describes a first-order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\). In first-order reactions, the rate of reaction is proportional to the concentration of the reactant.

Formula for First-order Reactions

For a first-order reaction, we use the formula: \[k = \frac{1}{t} \ln \left( \frac{a}{a-x} \right)\]where \( k \) is the rate constant, \( t \) is the time, \( a \) is the initial concentration of \( \mathrm{A} \), and \( x \) is the change in concentration of \( \mathrm{A} \).

Input Known Values

From the problem, after 60 minutes, the ratio \( \frac{a}{a-x} \) is 8. The initial concentration \( a \) is given as 0.1 M, and \( t = 60 \) minutes. Substitute these values into the formula:\[k = \frac{1}{60} \ln(8)\]

Calculate the Rate Constant

Calculate the natural logarithm and the rate constant:\[\ln(8) = 2.079\]\[k = \frac{1}{60} \times 2.079 = 0.03465 \text{ min}^{-1}\]

Calculate the Rate of Reaction

The rate of a first-order reaction in terms of concentration change per time is given:\[\text{Rate} = k \times a\]Substitute the value of \( k \) and \( a \):\[\text{Rate} = 0.03465 \text{ min}^{-1} \times 0.1 = 0.003465 \text{ mol L}^{-1} \text{ min}^{-1}\]

Select the Correct Option

Compare calculated rate with given options.The closest match is \(3.466 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}\).Thus, the answer is (b).

Key concepts

Rate Constant

In the study of chemical kinetics, the rate constant \(k\) plays an essential role. It links the speed of a reaction to the concentration of reactants. For first-order reactions, the rate constant is pivotal because it quantifies how quickly a reaction progresses independently of initial concentrations. Knowing \(k\) allows you to predict how fast a reactant will be converted into a product.

In the context of the problem, the rate constant is calculated using the formula for first-order reactions:

• \[k = \frac{1}{t} \ln \left( \frac{a}{a-x} \right)\]

Here, \(t\) is the reaction time, and \(\ln\) signifies the natural logarithm of the concentration ratio. By plugging in known values, we calculated \(k\) to be \(0.03465 \text{ min}^{-1}\). This rate constant is crucial because it helps determine the reaction's speed under given conditions.

Reaction Order

The order of a reaction describes how the rate is affected by the concentration of reactants. In a first-order reaction, the rate is directly proportional to the concentration of one reactant. This implies that if you double the concentration, the rate also doubles.

Understanding reaction order is fundamental because it affects the entire approach to solving kinetics problems. It determines how we set up rate equations and interpret experimental data. For first-order reactions, the formula \[k = \frac{1}{t} \ln \left( \frac{a}{a-x} \right)\] reflects this direct proportionality.
Knowing the reaction order allows scientists and engineers to predict how changing conditions can speed up or slow down a reaction. For example, in industrial processes, understanding reaction order ensures efficiency and safety, guiding decisions about the concentration of chemicals involved.

Concentration Change

Concentration change in the context of chemical reactions is the shift in the amount of reactant as the reaction progresses towards product formation. This change is crucial in predicting how the reaction proceeds over time.

In first-order reactions, the concentration change of a reactant \(A\) ("a \(\longrightarrow\) P") is straightforward to calculate. Initially, a molarity \(a\) is given. As the reaction progresses, part of this initial concentration is converted to product, represented as \(a-x\) where \(x\) is the amount reacted.
The problem demonstrates how this change relates to the reaction rate. Given a ratio of \(a/(a-x) = 8\) after 60 minutes, we establish how much of \(A\) has been used to form \(P\). The mathematical application here is significant as it directly leads to understanding reaction progress and determining reaction rates using the rate constant. As concentration decreases, the rate equation \(\text{Rate} = k \times a\) tells us how \(A\) is consumed over time, influencing practical decisions in fields from pharmaceuticals to chemical manufacturing.