The following data were measured for the reaction bf3(g)+nh3(g)→f3bnh3(g): experiment [bf3](m) [nh3](m) initial rate (m/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.1193 5 0.175 0.100 0.0596 part a what is the rate law for the reaction?
a. rate=k[bf3]2[nh3]
b. rate=k[bf3][nh3]
c. rate=k[bf3][nh3]2
d. rate=k[bf3]2[nh3]2

The right answer is "B": rate = k[BF3][NH3]

Explanation:

Let´s begin writting the generic rate law

rate = k[BF3]ᵃ[NH3]ᵇ

I will rewrite the data provided for clarification:

Experiment      [BF₃] (M)      [NH₃] (M)      initial rate (M/s)

1                         0.250           0.250           0.2130

2                        0.250            0.125            0.106

3                         0.200            0.100           0.0682

4                         0.350            0.100            0.1193

5                          0.175            0.100            0.0596

For experiments 1 and 2 the reaction rate laws can be written as follows:

v₁ = k([BF3]₁)ᵃ([NH3]₁)ᵇ (The subscripts indicate the number of experiment)

v₂ = k([BF3]₂)ᵃ([NH3]₂)ᵇ

we can write [BF3]₁ and [NH3]₁ in terms of [BF3]₂ and [NH3]₂ respectively:

[BF3]₁ = [BF3]₂

[NH3]₁ /  [NH3]₂ = 0.250 M / 0. 125 M = 2. Then:

[NH3]₁ = 2[NH3]₂

Then, replacing in v₁:

v₁ = k([BF3]₂)ᵃ (2[NH3]₂)ᵇ

If we divide v₁ / v₂:

v₁ / v₂ = k([BF3]₂)ᵃ (2[NH3]₂)ᵇ / k([BF3]₂)ᵃ ([NH3]₂)ᵇ  (distributing the exponent):

v₁ / v₂ = k([BF3]₂)ᵃ 2ᵇ([NH3]₂)ᵇ / k([BF3]₂)ᵃ ([NH3]₂)ᵇ (cancelating terms that are equal)

v₁ / v₂ = 2ᵇ (applying ln in both sides of  the equation)

ln(v₁ / v₂) = ln 2ᵇ (applying logarithmic property)

ln(v₁ / v₂) = b ln 2

ln(v₁ / v₂) / ln 2 = b

b = ln ((0.2130 M/s) / (0.106 M/s)) / ln 2 = 1

The same could be done using experiment 4 and 5 ( I choose them because [BF₃]₄ = 2 [BF₃]₅ and it makes calculation easier, but you can choose every combination you want):

v₄ = k([BF3]₄)ᵃ([NH3]₄)ᵇ

v₅ = k([BF3]₅)ᵃ([NH3]₅)ᵇ

[BF3]₄ = 2 [BF3]₅ and [NH3]₄ = [NH3]₅

Replacing in the equation v₄:

v₄ = k(2[BF3]₅)ᵃ ([NH3]₅)ᵇ = k 2ᵃ ([BF3]₅)ᵃ ([NH3]₅)ᵇ

v₅ = k([BF3]₅)ᵃ([NH3]₅)ᵇ

v₄ / v₅ = k 2ᵃ ([BF3]₅)ᵃ ([NH3]₅)ᵇ / k([BF3]₅)ᵃ([NH3]₅)ᵇ

v₄ / v₅ = 2ᵃ

ln (v₄ / v₅) = a ln 2

ln (v₄ / v₅) / ln 2 = a

ln ((0.1193 M/s) / (0.0596 M/s)) / ln 2 = a = 1

The right answer is "B".

Have a nice day!

Well reflection in chem is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated

Explanation: