Question

The following initial rate data apply to the raction

F2(g) + 2Cl2O(g) ---> 2FClO2(g) +Cl2(g)
Expt. [F2] (M) [Cl2O] (M) Intitial rate (M/s)
1 0.05 0.010 5 x 10^-4
2 0.05 0.040 2.0 x 10^-3
3 0.10 0.010 1.0 x 10^-3
Which of the following is the rate law (rate equation) for this reaction?
A. rate= k[F2]^2 [Cl2O]^4
B. rate= k[F2]^2 [Cl2O]
C. rate= k[F2] [Cl2O]
D. rate= k[F2] [Cl2O]^2
E. rate= k[F2]^2 [Cl2O]^2

Answer 1

C. rate = k[F₂] [Cl₂O]

Step-by-step explanation:

Based on the reaction, rate law can be obtained from the initial concentration of reactants thus:

rate = k[F₂]ᵃ [Cl₂O]ᵇ

Where the exponents a and b can be finded doing a experiment changing initial concentrations and seeing how a variation contribute in rate law.

If you analize experiments 1 and 2, the only change is [Cl₂O] (From 0.010 to 0.040, four times more) that changes its concentration in four times. This change produce rate law change from 5x10⁻⁴ to 2.0x10⁻³, also four times. That means the exponent b of [Cl₂O] is 1.

rate = k[F₂]ᵃ [Cl₂O]ᵇ

rate = k[F₂]ᵃ [Cl₂O]¹

Now, comparing experiments 1 and 3, the [F₂] change from 0.05 to 0.10, (Twice), and initial rate change from 5x10⁻⁴ to 1x10⁻³ (Also, twice). That means a = 1 and rate law is:

rate = k[F₂]¹ [Cl₂O]

rate = k[F₂] [Cl₂O]

Thus, right answer is:

C. rate = k[F₂] [Cl₂O]