t$$_{87.5}$$ is the time required for the reaction to undergo $$87.5\%$$ completion and t$$_{50}$$ is the time required for the reaction to undergo $$50\%$$ completion. The relation between t$$_{87.5}$$ and t$$_{50}$$ for a first order reaction is
t$$_{87.5}$$ = x $$\times$$ t$$_{50}$$
The value of x is _____. (Nearest integer)

For a first-order reaction, the time for a given percentage completion is $$t = \dfrac{2.303}{k}\log\dfrac{100}{100-p}$$, where $$p$$ is the percentage completed and $$k$$ is the rate constant.

For 50% completion: $$t_{50} = \dfrac{2.303}{k}\log\dfrac{100}{50} = \dfrac{2.303}{k}\log 2$$.

For 87.5% completion: $$t_{87.5} = \dfrac{2.303}{k}\log\dfrac{100}{12.5} = \dfrac{2.303}{k}\log 8 = \dfrac{2.303}{k}\cdot 3\log 2$$.

Taking the ratio: $$\dfrac{t_{87.5}}{t_{50}} = \dfrac{3\log 2}{\log 2} = 3$$. Therefore $$x = \boxed{3}$$. This also follows from the fact that 87.5% completion corresponds to 3 half-lives (the fraction remaining is $$1/8 = (1/2)^3$$), while 50% completion corresponds to 1 half-life.