Question

The ratio $t_{7/8}:t_{1/2}$ for the first order reaction is

• A. 3
• B. 5
• C. 2
• D. 7

Answer 1

Answer: (A)

$t_{7/8}=\frac{2.303}{K}\log \frac{a}{a-\frac{7}{8}a}=\frac{2.08}{K}$

$t_{1/2}=\frac{2.303}{K}\log \frac{a}{\left(a-\frac{a}{2}\right)}=\frac{0.693}{K}$

$\therefore \frac{t_{7/8}}{t_{1/2}}=3$