Answer: The rate constant for the reaction is [tex]3.96\times 10^{-3}min^{-1}[/tex]
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample = 559 min
a = let initial amount of the reactant = [tex]2.83\times 10^{-3}[/tex]
a - x = amount left after decay process = [tex]3.06\times 10^{-4}[/tex]
[tex]559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}[/tex]
[tex]k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}[/tex]
[tex]k=3.96\times 10^{-3}min^{-1}[/tex]
The rate constant for the reaction is [tex]3.96\times 10^{-3}min^{-1}[/tex]
