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) \(3: 2\)

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.

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.

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.