Question

The rate constant for a certain reaction is k = 4.50×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 11.0 minutes?

Answer 1

Answer: The concentration of reactant after the given time is 0.0205 M

Explanation:

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,  

k = rate constant  = $4.50\times 10^{-3}s^{-1}$

t = time taken for decay process = 11.0 min = 660 s  (Conversion factor:  1 min = 60 s)

$[A_o]$ = initial amount of the reactant = 0.400 M

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

$4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}$

$[A]=0.0205M$

Hence, the concentration of reactant after the given time is 0.0205 M