Question

Given these data in a study on how the rate of a reaction was affected by the concentration of the reactants,

Experiment [A] [B] [C] Rate (mol L‑1 hr‑1 )
1 0.200 0.100 0.600 5.0
2 0.200 0.400 0.400 80.0
3 0.600 0.100 0.200 15.0
4 0.200 0.100 0.200 5.0
5 0.200 0.200 0.400 20.0
From this data, what is the numerical value of the rate constant, (k), for this reaction (value that would be found using the same units used in the data above)?
a. 2083
b. 694
c. 417
d. 2500
e. 83.3

Answer 1

d. 2500

Step-by-step explanation:

In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:

v = k [A]ᵃ [B]ᵇ [C]ⁿ

If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:

v = k [A]ᵃ [B]ᵇ [C]⁰

v = k [A]ᵃ [B]ᵇ×1

Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:

v = k [A]ᵃ [B]ᵇ

v = k [A]ᵃ [B]²

Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:

v = k [A]ᵃ [B]²

v = k [A] [B]²

Relpacing this equation in any experiment (Experiment 5, for example):

20.0 = k [0.200] [0.200]²

2500 = k

That means right answer is:

d. 2500