Question

The half-life periods of two first-order reactions are in the ratio \(3: 2\). If \(t_{1}\) is the time required for \(25 \%\) completion of the first reaction and \(t_{2}\) is the time required for \(75 \%\) completion of the second reaction, then the ratio, \(t_{1}: t_{2}\), is \((\log 3=0.48\), \(\log 2=0.3\) ) (a) \(3: 10\) (b) \(12: 25\) (c) \(3: 5\) (d) \(3: 2\)

Short answer

The ratio, t1:t2, is 3:2.

Step-by-step solution

Understand Half-Life in First-Order Reactions

For a first-order reaction, the half-life (t1/2) is the time required for exactly one-half of the entity to react. The half-life is given by the formula: t1/2 = ln(2)/k, where k is the rate constant. Since the half-life periods of the two reactions are in a ratio of 3:2, let us assume that the half-lives are 3x and 2x respectively.

Calculate Time for 25% Completion - First Reaction

To find t1, the time required for 25% completion of the first reaction, we use the first-order reaction formula, where concentration decreases to (1 - 0.25) = 0.75 of its initial value. The formula is: t1 = (ln(1/0.75))/k1. We know that t1/2 for the first reaction is 3x, so k1 = ln(2)/(3x). Substitute k1 in the equation to find t1 in terms of x.

Calculate Time for 75% Completion - Second Reaction

To find t2, the time required for 75% completion of the second reaction, the concentration decreases to (1 - 0.75) = 0.25 of its initial value. So t2 = (ln(1/0.25))/k2. The second reaction's half-life is 2x, which means k2 = ln(2)/(2x). Substitute k2 in the equation to find t2 in terms of x.

Find the Ratio t1:t2

We have t1 and t2 in terms of x and ln(2). To find the ratio t1:t2, divide t1's expression by t2's expression. Simplify the expression using given logarithmic values (ln(2) = 0.69). The final ratio gives the answer.

Choose the Correct Option

After simplifying, match the calculated ratio t1:t2 with the given options to choose the correct answer.

Key concepts

Rate Constant

When discussing the kinetic details of chemical reactions, the term \textbf{rate constant} is key. It's a proportionality constant in the rate equation that provides the speed at which a reaction proceeds. Specifically, for a first-order reaction, the rate constant 'k' is involved in determining the relationship between the time taken for a reaction to reach a certain completion percentage and the half-life of the reaction.

In our example, for each reaction, the half-life can be expressed via the formula:
\( t_{1/2} = \frac{\ln(2)}{k} \).
By knowing the half-life of the reaction, we can derive the rate constant 'k' which plays a pivotal role in predicting the time a reaction will take to reach a given percentage of completion. The half-life inversely depends on the rate constant, meaning that a higher 'k' indicates a faster reaction, resulting in a shorter half-life. This relationship is crucial in various applications, including pharmaceuticals where the rate at which a drug degrades or reacts can determine its shelf life and dosage.

Reaction Completion Time

The \textbf{reaction completion time} is a measure of how long it takes for a reaction to progress to a certain point. In the context of our exercise, we are investigating the time it takes for the first reaction to reach 25% completion (\(t_1\)) and for the second reaction to reach 75% completion (\(t_2\)).

These points of completion are significant because they help us comprehend the reaction kinetics and the efficiency of the reaction process. To determine these times, we utilize a logarithmic expression derived from the integrated rate law for first-order reactions:
\( t = \frac{\ln(\frac{1}{1 - fraction\textunderscore complete})}{k} \).
It is important to apply this formula correctly and understand that for any given percentage of reaction completion, we do not simply use the completion percentage but rather use the remaining fraction of the reactant to perform the calculation. For instance, for 25% reaction completion, we use 0.75 (the remaining fraction) in the equation, not 0.25.

Logarithmic Calculations in Chemistry

Chemistry frequently employs \textbf{logarithmic calculations}, particularly in kinetics and equilibrium analysis. Logarithms help simplify multiplicative processes into additive ones, a feature that is especially handy when dealing with rates and constants of reactions.

Understanding the logarithm rules is essential when analyzing kinetic data. For example, we commonly see natural logarithms (ln) in these formulas because rates of reactions are often expressed in terms of e (Euler's number). The calculations in our exercise required the use of logarithmic identities to solve for the time ratios (\(t_1:t_2\)) with given logarithm values of common numbers.
Correctly applying logarithms to solve rate equations or equilibrium constants can transform nonlinear relationships into linear ones, greatly facilitating the interpretation and comparison of chemical reaction behaviors. Grasping the relationships between half-life, rate constants, and reaction completion will further solidify a student's understanding of reaction kinetics.