For a first-order reaction, what is the ratio between the time taken to complete three fourth of the reaction and the time taken to complete half of the reaction?
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 2:1.
For a first order reaction, t = \(\frac{2.303}{k}\)log\(\frac{a}{a-x}\)
For \(\frac{3}{4}\) of the reaction to occur, t = t3/4, (a - x) = a - \(\frac{3a}{4}\) = \(\frac{a}{4}\)
∴ t3/4 = \(\frac{2.303}{k}\)log4
For half of a reaction to occur, t = t1/2, (a - x) = a - \(\frac{a}{2}\) = \(\frac{a}{2}\)
∴ t1/2 = \(\frac{2.303}{k}\)log\(\frac{a}{\frac{a}{2}}\) = \(\frac{2.303}{k}\)log2
Hence, \(\frac{t_{3/4}}{t_{1/4}}\) = \(\frac{log4}{log2}\) = \(\frac{0.06021}{0.3010}\) = 2