Question

Use the following information to answer the following 3 questions:

2 Ti (NO3)2 (aq) + Ag (OH)4 (aq) <---> 2 Ti (OH)2 (aq) + Ag (NO3)4 (aq)

Trial    Ti(NO3)4, M    Ag(OH)4 M    R, M/s

Trial 1:    0.0250 0.0442 3.4 x 10^-8

Trial 2:    0.0133 0.0224    4.9 x10^-9

Trial 3:    0.0133 0.0442    9.7 x 10^-9

a) determine the rate law and calculate the rate constant for the forward reaction:

Reaction forward: 2 Ti(NO3)2 (aq) + Ag(OH)4 (aq) <--> 2 Ti(OH)2 (aq) + Ag(NO3)4 (aq)

b) if the initial concentration of Ag(OH)4 is 0.0250 M and it's half life is 354.8 minutes. Determine how long it will take for the concentration of Ag(OH)4 to decrease to 37.15%

Answer 1

a)

The rate law expression for the given reaction is as

Rate = k [Ti(NO₃)₂]^m[Ag(OH₄]ⁿ -------- (i)

For Trial 1, 2, and 3;

3.4 x 10^-8 = k [0.0250]^m[0.0442]ⁿ -------- (i)

4.9 x 10^-9 = k [0.0133]^m[0.0224]ⁿ -------- (ii)

9.7 x 10^-9 = k [0.0133]^m[0.0442]ⁿ -------- (iii)

From equation (i) & (iii), we get

(3.4 x 10^-8)/(9.7 x 10^-9) = [0.0250/0.0133]^m

3.505 = [1.88]^m
or

log(3.505) = m log(1.88)

m = log(3.505)/ log(1.88) = 2

m = 2

From equation (ii) & (iii), we get

(4.9 x 10^-9)/(9.7 x 10^-9) = [0.0224/0.0442]ⁿ

0.51 = [0.51]ⁿ

n = 1

Therefore, the rate law for the given reaction is as

Rate = k [Ti(NO₃)₂][Ag(OH₄]²

Calculate the rate constant using above equation,

3.4 x 10^-8 = k [0.0250]²[0.0442]¹

k = (3.4 x 10^-8)/(2.76 x10^-5) = 1.23 x 10^-3

rate constant, k = 1.23 x 10^-3

b)

We know that

N_(t) = N₍₀₎ e^{- kt}      

k = 0.693/t_1/2

k = 0.693/354.8 = 1.0 x 10^-3 minute^-1

N_(t) =(0.0250 x 37.15)/100 = 0.0093 M

N₍₀₎ = 0.0250 M

Now,

(0.0093) = (0.0250) e^{(- 1.0 x 10^-3 x t)}    

e^{(- 1.0 x 10^-3 x t)}    = (0.0093)/ (0.0250) = 0.372

(- 1.0 x 10^-3 x t) ln(e) = ln(0.372) = -0.988

t = (-0.988)/( (-1.0 x 10^-3) = 988.9 minutes

t = 988.9 minutes