|t/sec||1242 sec||2745 sec||4546 sec|
Here [ A]0 is proportional to the alkali consumed between t = &&& and t = 339 sec, i.e., [A]0 = (39.81 mL – 26.34 mL) = 13.47 mL [A]t is proportional to the alkali consumed at &&& minus the alkali consmed at the specified time. Thus,
[A], at 1242 sec = 39.81 mL – 27.80 mL = 12.01 mL
[A], at 2745 sec = 39.81 mL – 29.70 mL = 10.11 mL
[A], at 4546 sec = 39.81 mL – 31.81 mL = 8.00 mL
Substituting the values of t, [A]0 and [A]t in the first-order rate equation
we get the following values of k
Since the three values of k derived from first-order reaction are close to being identical, hydrolysis of methyl acetate is a first order reaction.
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.