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)

For a first-order reaction we use the integrated rate-law written with common (base 10) logarithms:

$$k \;=\; \frac{2.303}{t}\,\log\!\left(\frac{[A]_0}{[A]}\right)$$

Here $$k$$ is the rate constant, $$t$$ is the elapsed time, $$[A]_0$$ is the initial concentration and $$[A]$$ is the concentration remaining after time $$t$$.

We are first told that 75 % of the reactant is consumed in 90 minutes. 75 % consumed means 25 % left, so

$$\frac{[A]}{[A]_0}=0.25 \quad\Longrightarrow\quad \frac{[A]_0}{[A]}=\frac{1}{0.25}=4.$$

Substituting these numbers into the formula gives

$$k = \frac{2.303}{90}\,\log 4.$$

Using the given value $$\log 2 = 0.30$$ we obtain $$\log 4 = 2\,\log 2 = 2 \times 0.30 = 0.60$$, so

$$k = \frac{2.303}{90} \times 0.60.$$

Now we are asked to find the time required for 60 % of the reaction to be completed. 60 % consumed means 40 % left, hence

$$\frac{[A]}{[A]_0}=0.40 \quad\Longrightarrow\quad \frac{[A]_0}{[A]}=\frac{1}{0.40}=2.5.$$

Applying the same first-order expression for this new situation, we write

$$k = \frac{2.303}{t_{60\%}}\;\log 2.5,$$

where $$t_{60\%}$$ is the unknown time for 60 % completion. Solving for $$t_{60\%}$$ gives

$$t_{60\%} = \frac{2.303}{k}\,\log 2.5.$$

We already have $$k = \dfrac{2.303}{90} \times 0.60$$, so substituting this value of $$k$$ into the expression for $$t_{60\%}$$ yields

$$t_{60\%} = \frac{2.303}{\dfrac{2.303}{90}\times 0.60}\;\log 2.5.$$

The factor $$2.303$$ cancels out from numerator and denominator, leaving

$$t_{60\%} = \frac{90}{0.60}\,\log 2.5.$$

Using the given value $$\log 2.5 = 0.40$$, we have

$$t_{60\%} = \frac{90}{0.60} \times 0.40.$$

Carrying out the multiplication and division step by step:

$$\frac{90}{0.60} = 150,$$

so

$$t_{60\%} = 150 \times 0.40 = 60.$$

Therefore, approximately 60 minutes are required for 60 % completion of the same first-order reaction.

So, the answer is $$60 \text{ minutes}$$.