Question

Consider the he kinetics data below. Note that two reactions were studied. Use the information to calculate the EQUILIBRIUM CONSTANT for

            2 NOCl (g) -> 2NO(g) + Cl₂  

                                                                        [NO]               [Cl₂]                initial rate

                                                                        0.10M             0.10M             0.54M/s

                                                                        0.10M             0.20M             1.05M/s

2NO(g) + Cl₂(g) -> 2NOCl(g)                    0.10M             0.30M             1.62M/s

                                                                        0.10M             0.40M             2.16M/s

                                                                        0.20M             0.50M             10.8 M/s

                                                                        0.30M             0.50M             24.3 M/s

                                                                        0.40M             0.50M             43.2 M/s

[NOCl]            time

5.00M             0s

4.55M             5.0 x 10⁵s

2NOCl(g) -> 2NO(g) +   Cl₂(g)                   4.174M           1.0 x 10⁶s

3.855M           1.5 x 10⁶s

3.582M           2.0 x 10⁶s

Answer 1

first we have to find out rate law for forward reaction and backward reaction

forward reaction rate law & rate constant :

rate = Kf [NO]^x [Cl2]^y

from data

0.54 = Kf [0.1]^x [0.1]^y ---------------------------> 1

1.05 = Kf [0.1]^x [0.2]^y---------------------------> 2

10.8 = Kf [0.2]^x [0.5]^y---------------------------> 3

43.2 = Kf [0.4]^x [0.5]^y---------------------------> 4

by solving 1 and 2 we get y = 1

by solving 3,4 we get x =2

rate = Kf [NO]² [Cl2]¹

0.54 = Kf (0.1)^2 (0.1)

Kf = 540 ------------------------------forward reaction rate constant

backward reaction rate law & rate constant :

rate = k [NOCl]^x

use data

first we have try for first order rate constant

K= (2.303/t )(log Ao/A)

K = rate constant

t = time

A = initial concentration

Ao = concentration after time 't '

data(2)

K = (2.303 / 5 x10^5 ) x log (5/4.55)

    = 1.8 x 10^-7 sec-1

data (3)

K = (2.303 /10^6 ) x log (5/4.174)

    = 1.8 x 10^-7 sec-1

data (4)

K =(2.303 / 1.5 x10^6 ) x log (5/3.855)

   = 1.78 x 10^-7 sec-1

in all cases rate constant is constat. so it is first order reaction

rate constant for backward reaction

Kb = 1.78 x 10^-7

now equilibrium constant Kc = Kf/Kb

       Kc = 540 / 1.78 x 10^-7

            =3 x 10^9

equilibrium constant Kc = 3 x 10^9