Find the rate law, the individual and overall reaction orders, and the average value of k for the reaction h2 + i2 latex: \longrightarrow⟶ 2hi, using the following data at 450°c: exp initial rate (m/s) initial h2 (m) initial i2 (m) 1 1.9x10-23 0.0113 0.0011 2 1.1x10-22 0.0220 0.0033 3 9.3x10-23 0.0550 0.0011 4 1.9x10-22 0.0220 0.0056 order of h2 (enter a number) order of i2 (enter a number) overall order of the reaction (enter a number) value of k (multiply the value you get by 1e20 to avoid having to enter e notation, enter to no decimal places)

Order of H₂ = 1

Order of I₂ = 1

Overall order of reaction = 2

k value = 1.5 × 10 ⁻¹⁸ M⁻¹ · s⁻¹

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

no, the red giant is the hottest with 2.00.000.000 degrees celsius. while the white dwarf only 25,000 degrees celsius.

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