Answer: 0.33 M
Explanation:
As the order of the reaction is one, 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]3.4\times 10^{-3}\text{s}^{-1}[/tex]
t = age of sample  = 2.0 min = [tex]2.0\times 60s=120s[/tex]
a = let initial amount of the reactant  = 0.50 M
a - x = amount left after decay process  = ?
Now put all the given values in above equation, we get
[tex]120=\frac{2.303}{3.4\times 10^{-3}}\log\frac{0.50}{(a-x)}[/tex]
[tex]\log\frac{0.50}{(a-x)}=0.18[/tex]
[tex]\frac{0.50}{(a-x)}=1.5[/tex]
[tex]{(a-x)}=0.33M[/tex]
Thus the concentration of [tex]N_2O[/tex] remaining after 2.0 min is 0.33 M
