Question

Question 12 What is the concentration of reactant [A] at t = 15.00 s if A is 1st order, k = 0.428 s1 and the initial concentration of [A] = 8.30 M? Question 13. At what time (s) will [A] = 5.00 M if A is 2nd order, k = 0.428 M*s, and the initial concentration of [A] = 8.30 M? Question 14 Given the following rate law, find the half-life of the reactant o, if [0] = 0.322 M and k = 9.012x10 Ms.

Answer 1

Answer:

(12) For first-order reaction rate law is

ln[A]=-kt + ln[A]0

where [A]0=initial concentration of A,

k=rate constant, [A]= concentration of A at time t.

Given [A]0=8.30 M, k=0.428 s^-1, [A]=? at time t=15 s.

ln[A]=(- 0.428 s^-1 x 15 s) + ln[8.30]

ln[A]=-4.3

[A]=e^-4.3=0.0135 M.

[A]=0.0135 M.

(13) For first-order reaction rate law is

1/[A]=kt + 1/[A]0

where [A]0=initial concentration of A,

k=rate constant, [A]= concentration of A at time t.

Given [A]0=8.3 M, k=0.428 M^-1 s^-1, [A]=5 M at t=?

1/(5M)=(0.428 M^-1 s^-1 x t) + 1/(8.3 M)

(0.428 M^-1 s^-1 x t)=0.0795 M^-1

t=(0.0795 M^-1)/(0.428 M^-1 s^-1)

t=0.1858 s.

(14) Given rate constant, k=9.012 x 10^-3 M s^-1 and [O3]0=0.322 M

From the units of rate constant, the order od reaction is zero order.

Therefore half-life for the zero-order reaction is

t_1/2=[A]0/2k

where [A]0=initial concentration and t_1/2=half life.

Therefore here t_1/2=[O3]0/2k=(0.322 M)/(2 x 9.012 x 10^-3 M s^-1)

t_1/2=17.865 s.

Please let me know if you have any doubt. Thanks