Exercise 5 | Chemical Kinetics: Rate Laws

Here is a reaction: 2 NO(g) + Cl2(g) → 2 NOCl(g)

We measure the rate of reaction of 3 experiments with different concentration.

Determine the rate constant of the reaction.


Exp1: [NO] = 0.030 M; [Cl2] = 0.010 M; rate = 0.020 M.s-1

Exp2: [NO] = 0.015 M; [Cl2] = 0.010 M; rate = 0.0050 M.s-1

Exp3: [NO] = 0.015 M; [Cl2] = 0.040 M; rate = 0.020 M.s-1

Rate = k [NO]α [Cl2]β

 

Let’s determine α:

[NO]exp2 = 0.5 x [NO]exp1
[Cl2]exp2 = [Cl2]exp1
rateexp2 = 0.25 x rateexp1

rateexp2 = 0.25 x rateexp1
⇒ k [NO]exp2α [Cl2]exp2β = 0.25 k [NO]exp1α [Cl2]exp1β
⇒ k 0.5α x [NO]exp1 α [Cl2]exp1β = 0.25 k [NO]exp1α [Cl2]exp1β
⇒ 0.5α = 0.25
⇒ α = 2

 

Let’s determine β:

[NO]exp3 = [NO]exp2
[Cl2]exp3 = 4 [Cl2]exp2
rateexp3 = 4 x rateexp2

rateexp3 = 4 x rateexp2
⇒ k [NO]exp3α [Cl2]exp3β = 4 k [NO]exp2α [Cl2]exp2β
⇒ k [NO]exp2 α 4 β [Cl2]exp2β = 4 k [NO]exp2α [Cl2]exp2β
⇒ 4 β = 4
⇒ β = 1

 

Rate = k [NO]2 [Cl2]

 

Rateexp1 = k [NO]exp12 [Cl2]exp1

k = Rateexp1[NO]exp12 [Cl2]exp1 = 0.0200.0302 × 0.010 = 2.2 x 103 M-2.s-1