2. Cyclopropane, C3H6, is a gas used as a general anesthetic. It undergoes a slow molecular rearrangement to propylene. At a certain temperature, the following data were obtained relating concentration and rate.
Initial Concentration Initial Rate of Formation
(mol/L) of Propylene (mol/L•s)
[C3H6]
0.050 2.95 x 10^-5
0.100 5.90 x 10^-5
0.150 8.85 x 10^-5
What is the rate law for the rxn? What is the value of k?
To determine the rate law for the reaction and the value of k, we need to analyze the given data and determine how the rate of formation of propylene (C3H6) changes with respect to the concentration of cyclopropane (C3H6).
From the given data, we can see that the concentration of cyclopropane doubles from 0.050 to 0.100 mol/L, and the rate of formation of propylene also doubles from 2.95 x 10^-5 to 5.90 x 10^-5 mol/L•s. This suggests that the rate of the reaction is directly proportional to the concentration of cyclopropane, or mathematically:
Rate = k[C3H6]^n
Where "k" is the rate constant and "[C3H6]" is the concentration of cyclopropane.
To determine the value of "n" in the rate law, we can compare the change in rate with the change in concentration for two sets of data points. Let's compare the first two data points:
Rate1 / Rate2 = (0.050 / 0.100)^n
(2.95 x 10^-5) / (5.90 x 10^-5) = (0.050 / 0.100)^n
0.5 = 0.5^n
Taking the logarithm of both sides:
log(0.5) = log(0.5^n)
-0.301 = n * log(0.5)
n = (-0.301) / (log(0.5))
Using a calculator, we find that n ≈ -1.
Now that we have determined the value of "n" in the rate law, we can substitute it back into the rate law equation to find the complete rate law:
Rate = k[C3H6]^(-1)
Finally, we need to find the value of the rate constant "k." To do this, we can choose any set of concentration and rate values from the given data and substitute them into the rate law equation. Let's use the first data point:
(2.95 x 10^-5) = k * (0.050)^(-1)
k = (2.95 x 10^-5) / (0.050)
k ≈ 5.90 x 10^-4 mol/L•s
Therefore, the rate law for the reaction is Rate = k[C3H6]^(-1), and the value of k is approximately 5.90 x 10^-4 mol/L•s.