Problem 209 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{t}_{1}\) can be taken as the time taken for the concentration of a reactant to drop to \(\frac{3}{4}\) of its initial value. If the rate constant for a first order reaction is then \(\mathrm{t}_{1 / 4}\) can be written as (a) \(0.1 / k\) (b) \(0.29 / k\) (c) \(0.69 / k\) (d) \(0.75 / k\)

Step-by-Step Solution

Answer

The correct answer is (b) \(0.29/k\).

Step 1: Understand First Order Reactions

In a first-order reaction, the rate of the reaction depends linearly on the concentration of only one reactant. The rate law is given by \( \text{Rate} = k[A] \), where \([A]\) is the concentration of the reactant and \(k\) is the rate constant.

Step 2: Use First Order Reaction Formula

The formula for a first-order reaction concentration over time is \([A] = [A]_0 e^{-kt}\). We need the time when the concentration \([A]\) becomes \( \frac{3}{4} [A]_0 \).

Step 3: Solve for \( t_1 \)

Set the equation as \( \frac{3}{4}[A]_0 = [A]_0 e^{-kt_1} \). Divide both sides by \([A]_0\), resulting in \( \frac{3}{4} = e^{-kt_1} \).

Step 4: Use Natural Logarithm

Take the natural logarithm of both sides: \(\ln(\frac{3}{4}) = -kt_1\). This simplifies to \(t_1 = -\frac{\ln(\frac{3}{4})}{k}\).

Step 5: Numerical Calculation for \( t_1 \)

Calculate \( \ln(\frac{3}{4}) \), which is \( \ln(1) - \ln(\frac{4}{3}) \). \( \ln(\frac{4}{3}) \approx 0.2877 \), so \(-\ln(\frac{4}{3}) \approx -0.2877\). Therefore, \( t_1 \approx \frac{0.2877}{k} \).

Step 6: Conclusion and Option Match

Comparing \(t_1 = \frac{0.2877}{k}\) with the options, it most closely matches option (b) \(0.29/k\).

Key Concepts

Rate Constant

The rate constant, denoted by \(k\), is a crucial element in chemical reactions, specifically first-order reactions. In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. This means that the reaction rate is dictated by the change in concentration of the reactant. The formula is \(\text{Rate} = k[A]\), where \([A]\) represents the concentration of the reactant.

• For first-order reactions, the unit of the rate constant \(k\) is \(\text{s}^{-1}\), which stands for per second.
• The rate constant provides essential information about the speed of the reaction.
• The larger the \(k\), the faster the reaction occurs.
• Since \(k\) is specific to each reaction and condition, factors such as temperature can influence its value.

The rate constant aids in determining how long a reaction will take to reach a certain concentration level, which is why it plays a crucial role when calculating concentration over time.

Concentration Over Time

In a first-order reaction, understanding how concentration changes over time helps in predicting the behavior of the reaction. The concentration of the reactant decreases exponentially with respect to time. This property is captured by the equation for concentration over time: \[ [A] = [A]_0 e^{-kt} \] where:- \([A]\) is the concentration of the reactant at time \(t\).- \([A]_0\) is the initial concentration of the reactant.- \(e\) is the base of the natural logarithm.- \(k\) is the rate constant.

• This formula allows you to compute the concentration at any given time during the reaction process.
• It provides an understanding of how quickly the reactant concentration is reduced to a desired level.
• In our exercise, you need to find when the concentration becomes \( \frac{3}{4} \) of its initial value.

By setting up the equation \( \frac{3}{4}[A]_0 = [A]_0 e^{-kt_1} \) and solving for \(t_1\), you can calculate the time required for this change in concentration.

Natural Logarithm Calculation

Logarithms, specifically natural logarithms, are an invaluable tool in chemistry for solving exponential relations seen in concentration time equations. A natural logarithm, denoted \(\ln\), uses the constant \(e\) (approximately 2.718) as its base. In the context of a first-order reaction, natural logarithms are used to linearize the exponential decay equation:\[ \frac{3}{4} = e^{-kt_1} \]Taking the natural logarithm of both sides simplifies this equation:\[ \ln\left( \frac{3}{4} \right) = -kt_1 \]

• Natural logarithms convert multiplicative processes into additive ones, making them easier to solve algebraically.
• In our exercise, calculating \( \ln\left( \frac{3}{4} \right) \) leads to solving for \(t_1\), where \(t_1\) represents the time for the concentration to drop to \( \frac{3}{4} \) of its initial value.
• By solving \( t_1 = -\frac{\ln\left( \frac{3}{4} \right)}{k} \), we find that \(t_1\) approximately equals \( \frac{0.2877}{k} \), thus matching one of the given exercise options.

Natural logarithms hence play a crucial role in simplifying the complex relationships between variables in reaction equations.