Problem 71 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For a first-order reaction, \(t_{0.75}=1386 \mathrm{~s}\). Its specific reaction rate is (a) \(10^{-3} \mathrm{~s}^{-1}\) (b) \(10^{-2} \mathrm{~s}^{-1}\) (c) \(10^{-4} \mathrm{~s}^{-1}\) (d) \(10^{-5} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

\(k = 10^{-4} \mathrm{~s}^{-1}\)

Step 1: Understanding the Given Problem

We know that for a first-order reaction the time required to consume a certain fraction of reactant (in this case, 75%, which is the fraction left after the reaction) is related to the rate constant, k. The relationship is given by the formula: \( t = \frac{\ln(1/f)}{k} \), where \( f \) is the fraction of reactant remaining and \( t \) is the time. Here, \( t_{0.75} = 1386 \, s \) is the time at which 75% of the reactant remains unchanged.

Step 2: Substitute the Known Values into the Equation

We substitute \( t_{0.75} = 1386 \, s \) and \( f = 0.75 \) into the equation to find \( k \): \( 1386 = \frac{\ln(1/0.75)}{k} \). We now need to solve for \( k \).

Step 3: Calculate the Natural Logarithm Value

Calculate the value of \( \ln(1/0.75) = \ln(4/3) \) to find the numerator.

Step 4: Solve for the Specific Reaction Rate, k

Rearrange the equation to solve for \( k \), which gives \( k = \frac{\ln(4/3)}{1386} \). Next, we calculate this value to find the rate constant.

Step 5: Calculate the Specific Reaction Rate

Using a calculator, compute the value of \( k \) to determine the specific reaction rate. We calculate this to find which option (a, b, c, or d) corresponds with the calculated value.

Key Concepts

Chemical Kinetics

Chemical kinetics is the branch of physical chemistry that deals with understanding the rates of chemical reactions. It's essential for predicting how fast reactions proceed, which is crucial for a wide range of scientific and industrial processes.

Within chemical kinetics, the term 'first-order reaction' refers to a reaction where the rate is directly proportional to the concentration of one of the reactants. This implies that if you were to double the concentration of the reactant, the rate of reaction would also double. A characteristic feature of first-order reactions is their unique half-life, which remains constant regardless of the initial concentration. This concept of half-life and reaction orders helps chemists to design and control chemical processes by predicting the timing and yields of the reactions involved.

Reaction Rate

The reaction rate is a measure of how fast a chemical reaction occurs. For a first-order reaction, the rate can be expressed in terms of the change in concentration of a reactant, A, over time using the equation: \[ -\frac{d[A]}{dt} = k[A] \]where \( [A] \) is the concentration of A, \( t \) is time, and \( k \) is the first-order rate constant unique to the reaction. The minus sign indicates that the concentration of A decreases with time.

In our original exercise, rather than directly measuring the change in concentration, the time taken for a certain percentage of the reactant to remain (here 75%) is used to deduce the rate constant, illustrating a practical approach to studying reaction kinetics without detailed concentration data.

Rate Constant Calculation

The rate constant, represented by the symbol \( k \), is a coefficient that quantifies the rate of a chemical reaction. For a first-order reaction, the time required for a certain fraction of the reactant to remain can be used to calculate \( k \). The equation for a first-order reaction is \[ t = \frac{\ln(1/f)}{k} \], where \( f \) is the fraction of the original concentration of reactant remaining after time \( t \).

In the step-by-step solution provided, the fraction remaining is 0.75, and the time is 1386 seconds. By rearranging the formula and inserting the given values, one can solve for \( k \) to find which of the provided options corresponds to the calculated rate constant. It's a straightforward yet powerful tool for scientists to predict the behavior of reactions under different conditions.