Question

The following is known about the reaction below:

Chemical reaction: 2A-->2B

The order of A is second order

If you start out with 2.75 M of A, after 5 minutes you will have 2.00M

From these facts, determine:

a) the rate law constant, K

b) the first half life

c) from time 0, how long will it take to reach 25% of the original concentration of A

d) what is the concentration of A after 10 minutes?

Answer 1

according to second order reaction.

1/[A] = Kt + 1/[A₀]

[A] = 2.0 M

[A₀] = 2.75 M , t = 5.0 min

1/2 = K x 5 + 1/2.75

0.5 = 5K + 0.364

5K = 0.136

K = 2.72 x 10^-2 M^-1min^-1

b) t_1/2 = 1 / K[A₀]

t_1/2 = 1 / 2.72 x 10^-2 x 2.75

t_1/2 = 1 / 0.0748

t_1/2 = 13.37 min

c) initially = 2.75 M

25 % means

25 = X / 2.75 x 100

X = 0.6875 M

finally = 0.6875 M

1 / 0.6875 = 0.0272 x t + 1 /2.70

1.45 = 0.0272 t +0.370

0.0272 t = 1.08

t = 39.70 min

d) [A₀] = 2.70 M , t = 10 min , [A] = ?

1/[A] = 0.0272 x 10 + 1/2.70

1/[A] = 0.272 + 0.37

1/[A] = 0.642

[A] = 1.56 M