A first order reaction takes \(69.3\) minutes for \(50 \%\) completion. How much time will be needed for \(80 \%\) completion?

The rate constant, denoted as k, is a crucial number in chemical kinetics as it provides the speed at which a reaction proceeds. To compute the rate constant for a first-order reaction, we use the relationship:

\[\begin{equation}k = \frac{1}{t} \ln(\frac{1}{1 - x})\end{equation}\]

where

• t is the time taken for the reaction to reach a certain level of completion, and
• x is the fraction of reactant that has been converted at time t. For instance, if 50% of the reactant is consumed, x would be 0.5.

When t and x are known, as in the exercise problem at hand, the rate constant k can be calculated. In essence, it is a measure of the reaction's progress per unit time, and its unit is reciprocal of the time unit used, such as min^-1 or s^-1. Knowing k is pivotal because it allows us to predict the time required for different levels of reaction completion and is a fundamental step in solving many IIT-JEE chemistry problems.

The concept of half-life is widely used in describing first-order reactions. The half-life of a reaction, typically represented as t_1/2, is the time required for half of the reactant to be converted into products. For a first-order reaction, the half-life is constant and is given by the formula:

\[\begin{equation} t_{1/2} = \frac{0.693}{k}\end{equation}\]

where k is the rate constant. A prominent feature of first-order reactions is that their half-life does not depend on the initial concentration of reactants. This particularity makes calculations involving half-lives very convenient, as the same amount of time will pass for each successive half of the reactant to decompose, regardless of how much reactant is left.

The natural logarithm plays a significant role in the kinetics of chemical reactions, especially first-order reactions. It is an intrinsic part of the formula used to calculate the rate constant and, subsequently, the time taken for a certain level of reaction completion.

\[\begin{equation} t = \frac{1}{k} \ln(\frac{1}{1 - x})\end{equation}\]

In this equation, we take the natural logarithm of the inverse of one minus the fraction completed (x). This mathematical operation allows us to linearize the exponential decay of reactant concentration over time, which is a characteristic of first-order kinetics. Understanding how the natural logarithm is used in reaction kinetics is vital, not just for tackling textbook problems but also for interpreting experimental data in real-world scenarios.

In the context of IIT-JEE, one of India’s most challenging engineering entrance exams, chemistry problems often assess a student’s understanding of fundamental concepts, including reaction kinetics. First-order reaction calculations are a common topic in these competitive exams.

Applying the concepts of rate constant calculation, half-life, and the use of natural logarithms can help solve problems regarding reaction times and the extent of completion. Students are expected to seamlessly integrate these concepts to efficiently work through the problems, applying mathematical manipulations where necessary. For example, by utilizing the concept of half-life, a student might bypass the need to calculate the rate constant directly when the problem pertains to the time taken for half of the reactant to react. Familiarity and fluency with these topics enable students to tackle a wide array of IIT-JEE chemistry problems with confidence.