Consider the he kinetics data below. Note that two reactions were studied. Use t
ID: 879251 • Letter: C
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
Explanation / Answer
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