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

For a first-order reaction, the relation between the reaction rate constant (k) and time (t) for a given percentage of completion (p) is: <br/><br/> $\ln\left(\frac{1}{1-p}\right) = kt$ <br/><br/> For t₅₀ (50% completion), p = 0.5: <br/><br/> $\ln\left(\frac{1}{1-0.5}\right) = k\cdot t_{50}$<br/><br/> $\ln\left(\frac{1}{0.5}\right) = k\cdot t_{50}$<br/><br/> $\ln(2) = k\cdot t_{50}$ <br/><br/> For t₈₇.₅ (87.5% completion), p = 0.875: <br/><br/> $\ln\left(\frac{1}{1-0.875}\right) = k\cdot t_{87.5}$<br/><br/> $\ln\left(\frac{1}{0.125}\right) = k\cdot t_{87.5}$ $\ln(8) = k\cdot t_{87.5}$ <br/><br/> Now, we need to find the relationship between t₈₇.₅ and t₅₀: <br/><br/> $\frac{k\cdot t_{87.5}}{k\cdot t_{50}} = \frac{\ln(8)}{\ln(2)}$ <br/><br/> Since the k's cancel out, we have: <br/><br/> $\frac{t_{87.5}}{t_{50}} = \frac{\ln(8)}{\ln(2)}$ <br/><br/> Using the property of logarithms, we get: <br/><br/> $\frac{t_{87.5}}{t_{50}} = \frac{\ln(2^3)}{\ln(2)}$ <br/><br/> $\frac{t_{87.5}}{t_{50}} = 3$ <br/><br/> So, the value of x is 3.