Answer: The concentration of [tex]CH_3NC[/tex] will be [tex]1.56\times 10^{-2}M[/tex] after 416 seconds have passed.
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 = [tex]3.00\times 10^{-3}s^{-1}[/tex]
t = age of sample = ?
a = let initial amount of the reactant = [tex]5.45\times 10^{-2}M[/tex]
a - x = amount left after decay process = [tex]1.56\times 10^{-2}M[/tex]
[tex]t=\frac{2.303}{3.00\times 10^{-3}}\log\frac{5.45\times 10^{-2}}{1.56\times 10^{-2}}[/tex]
[tex]t=416s[/tex]
The concentration of [tex]CH_3NC[/tex] will be [tex]1.56\times 10^{-2}M[/tex] after 416 seconds have passed.