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!

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.