Question

A first order reaction has a rate constant of $2.303\times {10}^{-3}{s}^{-1}$. The time required for 40 g of this reactant to reduce to 10 g will be:
(Given that log2 =0.3010)

• A. 230.3 s
• B. 301 s
• C. 602 s
• D. 2000 s

Answer 1

The correct option is C 602 s

For a first order reaction,
$t=\frac{2.303}{k}log\frac{\left[A_0\right]}{\left[A_t\right]}$
$2.303\times {10}^{-3}=\frac{2.303}{t}log\frac{\left[40\right]}{\left[10\right]}$
$t=\frac{1}{{10}^{-3}}log{2}^2=\frac{2}{{10}^{-3}}log2=\frac{2}{{10}^{-3}}\times 0.3010=602s$