Answer:
The right answer is "B": rate = k[BF3][NH3]
Step-by-step 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!