Answer:
Order of H₂ = 1
Order of I₂ = 1
Overall order of reaction = 2
k value = 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹
Step-by-step explanation:
Experiment rate M/s [H₂] [I₂]
1 1.9 × 10⁻²³ 0.0113 0.0011
2 1.1 × 10 ⁻²² 0.0220 0.0033
3 9.3 × 10⁻²³ 0.0550 0.0011
4 1.9 × 10⁻²² 0.0220 0.0056
The generic rate law can be written as follows:
rate = k · [H₂]ᵃ · [I₂]ⁿ
Let´s take 2 experiments in which the concentration of one of the reactants is the same:
The rate law for experiment 1 will be:
1.9 × 10⁻²³ M/s = k · (0.0113 M)ᵃ · (0.0011 M)ⁿ
For experiment 3 :
9.3 × 10⁻²³ M/s = k · (0.0550 M)ᵃ · (0.0011 M)ⁿ
If we divide both rates:
rate 1 / rate 3 = 1.9 × 10⁻²³ M/s / 9.3 × 10⁻²³ M/s = 0.20
0.20 = (k · (0.0113 M)ᵃ · (0.0011 M)ⁿ) / (k · (0.0550 M)ᵃ · (0.0011 M)ⁿ)
0.20 = 0.20ᵃ
Applying ln:
ln 0.20 = a · ln0.20
a = 1
Now, let´s take the rate laws of experiment 4 and 2:
rate 4 / rate 2 = 1.9 × 10⁻²² / 1.1 × 10 ⁻²² = 1.7
1.7 = (k · (0.0220 M) · (0.0056 M)ⁿ) / (k · (0.0220 M) · (0.0033 M)ⁿ)
1.7 = 1.7ⁿ
n = 1
Then, the order of H₂ and I₂ is 1 and the order of the reaction is 1+1 = 2
To calculate the constant, we can solve the rate law equation for k for each experiment and then find the average value of k:
Experiment 1:
1.9 × 10⁻²³ = k · 0.0113 M · 0.0011 M
k = 1.9 × 10⁻²³M/s / 0.0113 M · 0.0011 M = 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹
Experiment 2
1.1 × 10 ⁻²² M/s = k · 0.0220 M · 0.0033 M
k = 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹
Experiment 3
9.3 × 10⁻²³ M/s = k · 0.0550 M · 0.0011 M
k = 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹
Experiment 4
1.9 × 10⁻²² M/s = k · 0.0220 M · 0.0056 M
k = 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹
The rate constant is 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹