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:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]4.50\times 10^{-3}s^{-1}[/tex]
t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)
[tex][A_o][/tex] = initial amount of the reactant = 0.400 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
[tex]4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}[/tex]
[tex][A]=0.0205M[/tex]
Hence, the concentration of reactant after the given time is 0.0205 M
