heating cyclopropane (C3H6) converts it to propene (CH2=CHCH3). The rate law is first order in cyclopopane. If the rate constant at a particular temp is 6.22x10^-4 s^1 and the concentration of cyclopropane is held at .0300 mol/L, what mass of propene is produced in 10.0 min in a volume of 2.50 Liters? I a, so lost on this. Any help is appreciated
ln(Ao/A) = akt
Ao = 0.03
a = 1
k = 6.22E-4
t = 10 min x (60s/min) - 600 s
Solve for (A) in mols/L
Plug in L to solve for mols
grams = mols x molar mass = ?
I got 517.5. I think I'm so off. Thanks anyway
Also im good at posting
To find the mass of propene produced in 10.0 minutes, we need to use the given information about the rate constant and the concentration of cyclopropane.
First, let's calculate the number of moles of cyclopropane present in the given volume of 2.50 liters. We can use the concentration of cyclopropane provided:
moles of cyclopropane = concentration of cyclopropane × volume
= 0.0300 mol/L × 2.50 L
Next, we can use the rate law to determine the rate of the reaction. Since the rate law is first order in cyclopropane, the rate of the reaction is given by:
rate = k × [cyclopropane]
Where:
rate = rate of reaction
k = rate constant
[cyclopropane] = concentration of cyclopropane
Given that the concentration of cyclopropane is held at 0.0300 mol/L, we can substitute these values into the rate equation:
rate = (6.22 × 10^-4 s^-1) × (0.0300 mol/L)
Now, let's convert the given time of 10.0 minutes into seconds:
time = 10.0 min × 60 s/min
Finally, we can determine the moles of propene produced using the rate and time values:
moles of propene = rate × time
To convert the moles of propene to mass, we'll need to use the molar mass of propene.
Step 1: Calculate the moles of cyclopropane:
moles of cyclopropane = 0.0300 mol/L × 2.50 L = 0.0750 mol
Step 2: Calculate the rate of the reaction:
rate = (6.22 × 10^-4 s^-1) × (0.0300 mol/L) = 1.87 × 10^-5 mol/L·s
Step 3: Convert the time to seconds:
time = 10.0 min × 60 s/min = 600 s
Step 4: Calculate the moles of propene:
moles of propene = rate × time = (1.87 × 10^-5 mol/L·s) × 600 s = 0.0112 mol
Step 5: Calculate the mass of propene using the molar mass of propene:
molar mass of propene (CH₂=CHCH₃) = (2 × atomic mass of carbon) + (4 × atomic mass of hydrogen)
= (2 × 12.01 g/mol) + (4 × 1.01 g/mol)
= 42.08 g/mol
mass of propene = moles of propene × molar mass of propene
= 0.0112 mol × 42.08 g/mol
= 0.470 g
Therefore, the mass of propene produced in 10.0 minutes in a volume of 2.50 liters is approximately 0.470 grams.