13.0k views
5 votes
The rate of decay of a chemical involved in a reaction that is second order (bimolecular) in one reactant A is given by: = k [AJ? -d[A/dt where k is the reaction rate coefficient. Derive an expression for the half-life of A in terms of k and the concentration of A at time t-0 (Ao)

1 Answer

4 votes

Step-by-step explanation:


2A\rightarrow products

According to mass action,


\textrm{rate}=-(\Delta[\textrm A])/(2\Delta t)=k[\textrm A]^2

Where, k is the rate constant

So,


(d[A])/(dt)=-k[A]^2

Integrating and applying limits,


\int_([A_t])^([A_0])(d[A])/([A]^2)=-\int_(0)^(t)kdt

we get:


(1)/([A]) = (1)/([A]_0)+kt

Where,


[A_t] is the concentration at time t


[A_0] is the initial concentration

Half life is the time when the concentration reduced to half.

So,
[A_t]=(1)/(2)* [A_0]

Applying in the equation as:


t_(1/2)=(1)/(k[A_o])

User Shali Liu
by
6.1k points