Question

A first order reaction has a specific reaction rate of 10^-2 s^-1. How much time will it take for 20 g of the reactant to reduce to 5 g?

• A. 238.6 s
• B. 138.6s
• C. 346.6s
• D. 693.0s

Answer 1

The correct option is B

138.6s

For a first order reaction,
Rate constant $\left(k\right)=\frac{2.303}{t}log\frac{a}{a-x}$
Where, a =initial concentration
a–x =concentration after time ‘t’
t =times in sec.
Given, a =20 g, a - x =5 g, $k={10}^{-2}$
$\therefore t=\frac{2.303}{{10}^{-2}}log\frac{20}{5}$
=138.6s
Alternatively,
Half –life for the first order reaction,
$t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{{10}^{-2}}=69.3s$
Two half –lives are required for the reduction of 20 g of reactant into 5 g.
$20g\stackrel{t_{\frac{1}{2}}}{\to }10g\stackrel{t_{\frac{1}{2}}}{\to }5g.$
$\therefore$ Total time $=2t_{\frac{1}{2}}=2\times 69.3=138.6s$