Question

For the reaction, a → b, the rate constant is 0.0208 m-1 sec-1. How long would it take for [a] to decrease from 0.100 to 0.0450 m?

Answer 1

First, assume the order of the given reaction is n, then the rate of reaction i.e. $\frac{dx}{dt}=k\times[A]^{n}$

where, dx is change in concentration of A in small time interval dt and k is rate constant.

According to units of rate constant, the reaction is of second order.

$\frac{1}{[A]_{t}}-\frac{1}{[A]_{o}} = kt$   (second order formula)

Put the values,

$\frac{1}{0.04590 m}-\frac{1}{0.100 m} =0.0208 m^{-1}s^{-1} \times t$  

 $22.23 m -10 m =0.0208 m^{-1}s^{-1} \times t$

$\frac{12.23 m}{0.0208 m^{-1}s^{-1}} = t$

t= 587.9 s

Hence, time taken is 587.9 s