Animated Solution for Chemistry - Chemical Kinetics: If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) ........... . (Take : $\log 2=0.30$; $\log 2.5=0.40$)

Decoding First-Order Kinetics

Welcome to a classic problem in Chemical Kinetics! When dealing with first-order reactions, the relationship between time, concentration, and the rate constant is governed by a beautiful logarithmic equation. Let's break down how to solve these problems efficiently without getting bogged down in messy calculations.

For any first-order reaction, the integrated rate law is our primary tool:

Here, $t$ is the time, $k$ is the rate constant, $[A{]}_0$ is the initial concentration, and $[A{]}_t$ is the concentration remaining at time $t$. The most common mistake students make is substituting the reacted amount instead of the remaining amount for $[A{]}_t$. Always remember: $[A{]}_t$ is what is left in the vessel!

The problem states that the reaction is $75\%$ complete in $90$ minutes. Let's assume our initial concentration $[A{]}_0$ is $100$. If $75\%$ has reacted, the remaining concentration $[A{]}_t$ is $100-75=25$.

Plugging this into our master equation:

This simplifies to $\log (4)$. The problem cleverly provides $\log 2=0.30$. Since $4=2^2$, we know that $\log (4)=2\log (2)=2\times 0.30=0.60$.

So, our first equation becomes:

Next, we need to find the time for $60\%$ completion. Using the same logic, if $60\%$ has reacted, the remaining concentration is $100-60=40$.

Setting up the equation for this new time $t_{60\%}$:

This simplifies to $\log (2.5)$. The problem generously gives us $\log 2.5=0.40$.

So, our second equation is:

Now, we could solve equation (1) for $k$ and then plug it into equation (2). But why do extra work? The most elegant way to solve this is to divide equation (2) by equation (1). This instantly cancels out the rate constant $k$ and the $2.303$ term!

Solving for $t_{60\%}$ is now a breeze:

By using the ratio method and the provided logarithmic values, we bypassed complex arithmetic and arrived at the exact answer swiftly. Always look for these mathematical shortcuts in competitive exams!