heating cyclopropane (C3H6) converts it to propane (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 propane is produced in 10.0 min in a volume of 2.50 Liters.
Can you please take me through this problem step by step, I have been trying different ways and now I am really confused.
Thanks C
Do you mean propEne? Do you mean k = 6.22E-4 s^-1?
rate = k*0.03 = mols/L*s
Then mols/L*s x (60 s/min) x 10 min x 2.5 L = mols in 2.5L in 10 min. Convert to grams. mols = grams/molar mass. You know molar mass and mols, solve for grams.
Ok I got .2799 moles, do i have to convert it to grams, how would I do that.
The problem asks for grams so I think you should make that conversion if you want to answer the question. For how you do it go back and read my response. I told you how to do it. Also you should punch those numbers into your calculator again. I didn't get that number. Also if your prof is picky about significant figures you will lose points because you have reported more than you are allowed.
Of course! I'll guide you through this problem step by step.
1. First, let's write down the chemical equation for the reaction:
Cyclopropane (C3H6) → Propane (CH2=CHCH3)
2. We know that the rate law is first order in cyclopropane. This means that the rate of the reaction is proportional to the concentration of cyclopropane raised to the power of 1.
3. The rate law equation can be written as:
rate = k[Cyclopropane]^1
Where:
- "rate" is the rate of the reaction
- "k" is the rate constant
- "[Cyclopropane]" represents the concentration of cyclopropane
4. We are given that the rate constant, k, at the particular temperature is 6.22x10^-4 s^-1.
5. The concentration of cyclopropane is held at 0.0300 mol/L.
6. We need to find the mass of propane produced in 10.0 minutes, which is equivalent to 600 seconds. The volume of the reaction vessel is 2.50 liters.
Now, let's calculate the mass of propane produced:
1. Calculate the number of moles of cyclopropane used by multiplying the concentration by the volume:
Moles of Cyclopropane = 0.0300 mol/L * 2.50 L = 0.0750 mol
2. Use the given rate constant to calculate the rate of the reaction using the rate law equation. Since the rate law is first order, the rate will be equal to the rate constant multiplied by the concentration of cyclopropane:
Rate = k * [Cyclopropane]
= (6.22x10^-4 s^-1) * (0.0300 mol/L)
= 1.866x10^-5 mol/(L*s)
3. Determine the number of moles of propane produced by using the stoichiometry of the reaction. Since the stoichiometric ratio between cyclopropane and propane is 1:1, the number of moles of propane produced will be the same as the number of moles of cyclopropane used:
Moles of Propane = 0.0750 mol
4. Convert the number of moles of propane to grams by multiplying by the molar mass of propane. The molar mass of propane (CH2=CHCH3) is 44.10 g/mol:
Mass of Propane = 0.0750 mol * 44.10 g/mol
= 3.3075 g
Therefore, the mass of propane produced in 10.0 minutes in a volume of 2.50 liters is 3.3075 grams.