Answer:
[tex]\large\boxed{\large\boxed{0.291M}}[/tex]
Explanation:
By definition one half-life is the time to reduce the initial concentration to half.
For a second order reaction the rate law equations are:
[tex]\dfrac{d[B]}{dt}=-k[B]^2[/tex]
[tex]\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt[/tex]
The half-life equation is:
[tex]t_{1/2}=\dfrac{1}{k[A]_0}[/tex]
Thus, substitute the rate constant [tex]1.30\times 10^{-3}M^{-1}\cdot s^{-1}[/tex] and the half-life time 224s to find [A]₀:
[tex]224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}[/tex]
[tex][A]_o=0.291M[/tex]
The concentration of reactant B after one half life is 1.715 M..
The term half life refers to the time take for half of the original concentration of the reactants to remain in the system.
We know that, for a second order reaction;
t1/2 = 1/k[A]o
Where;
t1/2 = half life
k = rate constant
[A]o = initial concentration
Hence;
224 = 1/1.30×10−3 * [A]o
224 * 1.30×10−3 * [A]o = 1
[A]o = 1/224 * 1.30×10−3
[A]o = 3.43 M
After one half life, about half of this concentration will remain, therefore, we will have 1.715 M.
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