Answer :
Answer:
a) The rate law is: v = k[NO]² [O₂]
b) The units are: M⁻² s⁻¹
c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹
d) The rate of disappearance of NO is 0.8 M/s
e) The rate of disappearance of O₂ is 0.4 M/s
Explanation:
The experimental rates obtained can be expressed as follows:
v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s
v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s
v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s
where:
k = rate constant
[NO]₁ = concentration of NO in experiment 1
[NO]₂ = concentration of NO in experiment 2
[NO]₃ = concentration of NO in experiment 3
[O₂]₁ = concentration of O₂ in experiment 1
[O₂]₂ = concentration of O₂ in experiment 2
[O₂]₃ = concentration of O₂ in experiment 3
a and b = order of the reaction for each reactive respectively.
We can see these equivalences:
[NO]₂ = 2[NO]₁
[O₂]₂ = [O₂]₁
[NO]₃ = [NO]₂
[O₂]₃ = 2[O₂]₂
So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :
v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ
If we rationalize v2/v1, we will have:
v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)
v2/v1 = 2ᵃ
ln(v2/v1) = a ln2
ln(v2/v1) / ln 2 = a
a = 2
(Please review the logarithmic properties if neccesary)
In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:
v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ
v3/v2 = 2ᵇ
ln(v3/v2) = b ln 2
ln(v3/v2) / ln 2 = b
b = 1
Then, the rate law for the reaction is:
v = k[NO]² [O₂]
Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:
M/s = k * M³
M/s * M⁻³ = k
M⁻² s⁻¹ = k
To obtain the value of k, we can solve this equation for every experiment:
k = v / [NO]² [O₂]
for experiment 1:
k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹
for experiment 2:
k = 7.11 x 10³ M⁻² s⁻¹
for experiment 3:
k = 7.12 x 10³ M⁻² s⁻¹
The average value of k is then:
(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = 7.11 x 10³ M⁻² s⁻¹
The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:
v = k [NO]² [O₂]
The rate of the reaction in terms of the disappearance of NO can be written this way:
v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)
where (Δ [NO] / Δt) is the rate of disappearance of NO.
Then, calculating v with the data provided by the problem:
v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s
Then, the rate of disappearance of NO will be:
2v = Δ [NO] / Δt = 0.8 M/s
The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = 0.4 M/s
With calculations:
v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).