For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is _________. (Integer answer)

For any first‐order reaction, the integrated rate equation relating the concentration at time $$t$$ to the initial concentration is written as

$$k\,t \;=\; 2.303\;\log\!\left(\dfrac{[\text{Initial}]}{[\text{Remaining at }t]}\right).$$

When we speak of percentage completion, it is convenient to keep the initial amount as $$100$$ units. If a reaction is $$x\%$$ complete, then $$x$$ units have reacted and $$100 - x$$ units still remain. Substituting this in the above expression we obtain

$$k\,t_x \;=\; 2.303\;\log\!\left(\dfrac{100}{100 - x}\right),$$

where $$t_x$$ is the time required for $$x\%$$ completion.

Now we calculate the individual times.

Time for 50 % completion

For $$x = 50$$ we have

$$k\,t_{50} \;=\; 2.303\;\log\!\left(\dfrac{100}{100 - 50}\right)$$

$$\;\;\;=\; 2.303\;\log\!\left(\dfrac{100}{50}\right)$$

$$\;\;\;=\; 2.303\;\log\!\bigl(2\bigr).$$

Hence

$$t_{50} \;=\; \dfrac{2.303}{k}\;\log 2.$$

Time for 75 % completion

For $$x = 75$$ the same formula gives

$$k\,t_{75} \;=\; 2.303\;\log\!\left(\dfrac{100}{100 - 75}\right)$$

$$\;\;\;=\; 2.303\;\log\!\left(\dfrac{100}{25}\right)$$

$$\;\;\;=\; 2.303\;\log\!\bigl(4\bigr).$$

Thus

$$t_{75} \;=\; \dfrac{2.303}{k}\;\log 4.$$

Forming the required ratio

We now divide $$t_{75}$$ by $$t_{50}$$:

$$\dfrac{t_{75}}{t_{50}} \;=\; \dfrac{\dfrac{2.303}{k}\,\log 4}{\dfrac{2.303}{k}\,\log 2}.$$

The common factor $$\dfrac{2.303}{k}$$ cancels out, leaving

$$\dfrac{t_{75}}{t_{50}} \;=\; \dfrac{\log 4}{\log 2}.$$

Using the logarithmic identity $$\log(4) = \log(2^2) = 2\,\log(2),$$ we substitute:

$$\dfrac{t_{75}}{t_{50}} \;=\; \dfrac{2\,\log 2}{\log 2} \;=\; 2.$$

So, the answer is $$2$$.