Question

It takes 1 h for a first order reaction to go to 50% completion. The total time required for the same reaction to reach 87.5% completion will be:

• A. 1.75 h
• B. 6.00 h
• C. 3.50 h
• D. 3.00 h

Answer 1

The correct option is D 3.00 h

$t_{1/2}=1hr{t}_{87.5\%}=\frac{2.303}{k}log\frac{a_o}{a_o-x}=\frac{2.303}{0.693}log\frac{a_o}{a_o-0.875a_o}=\frac{2.303}{0.693}log8log8=log{2}^3=3log2$
$=3.323\times 0.90=3$