Answer the following questions on the above curve for a first order reaction A → P.
(a) What is the relation between slope of this line and rate constant.
(b) (i) Calculate the rate constant of the above reaction if the slope is 2 x 10–4 s–1.
(ii) Derive the relationship between half life of a first order and its rate constant.
(a) Slope = – k
or k = – 2 x 10–4 mol–1 Ls–1.
(ii) Consider the following first order reaction
A → P
at t = 0 a 0
at t = t (a – x) x
Suppose a is the initial concentration of reactant. After time t, x gm mol lit–1 is changed to P. According to law of mass action the rate of reaction at time t is directly proportional to the concentration of A at that instant i.e., (a – x)
is the rate and K is the rate constant, after rearranging the above equation, we get
On integrating the above equation
When t = 0, x = 0 hence – In a = constant substituting this value of constant in eqn. (ii) we get
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.