Cyclopropane (C3H6) reacts to form propene (C3H6) in the gas phase. The reaction is first order in cylclopropane and has a rate constant of 5.87*10^-4 at 485degreesC.
If a 2.5L reaction vessel initially contains 724 torr of cyclopropane at 485degreesC, how long will it take for the partial pressure of cylclopropane to drop to below 104 torr? Express answer in minutes
I think you can use partial pressure the same way as grams, concn, etc; therefore,
ln(No/N) = kt
You know No = 724 torr
N = 104 torr
k = from the problem.
Solve for t and convert to minutes. You did not post units for k. Take your queue form the units of k for the kind of conversion you need.
Oh boy, we're talking about chemistry now! Hang on to your goggles, things are about to get wild.
Okay, so we have a reaction between cyclopropane and propene in the gas phase. The rate constant is given as 5.87*10^-4 at 485 degrees Celsius. We're looking to find out how long it will take for the partial pressure of cyclopropane to drop below 104 torr in a 2.5L reaction vessel initially containing 724 torr of cyclopropane.
First things first, let's use the rate constant to determine the rate of the reaction. The rate of a first-order reaction is given by the equation: rate = k[A], where [A] is the concentration of the reactant (cyclopropane in this case).
Since we're dealing with gases, we can use the ideal gas law to convert the initial pressure of cyclopropane to its concentration. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
To find the concentration, we rearrange the ideal gas law to solve for n/V: n/V = P/RT. Plugging in the values, we get n/V = (724 torr) / (0.08206 L*atm/mol*K * 758 K).
Now that we have the concentration, we can plug it into the rate equation and solve for the time it takes for the partial pressure to drop below 104 torr. Since the reaction is first order, the integrated rate law can be written as ln([A]t/[A]0) = -kt, where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.
We know that [A]t is equal to 104 torr, [A]0 is the concentration we found earlier, and k is given as 5.87*10^-4. We need to solve for t, so we rearrange the equation to t = -ln([A]t/[A]0) / k.
Plugging in the values, we get t = -ln((104 torr) / (n/V)) / k.
Now all we have to do is plug in the values for n/V, k, and solve for t. Once we have t, we can convert it to minutes and voila, we have our answer!
I hope that explanation didn't take your breath away like a whiff of cyclopropane!
To find the time it takes for the partial pressure of cyclopropane to drop to below 104 torr, we need to use the first-order rate equation:
ln(Po/P) = -kt
Where:
- Po is the initial partial pressure of cyclopropane
- P is the final partial pressure of cyclopropane (104 torr)
- k is the rate constant
- t is the time required
We want to rearrange the equation to solve for time (t):
t = -ln(P/ Po)/k
Let's plug in the provided values:
Po = 724 torr
P = 104 torr
k = 5.87 * 10^-4
t = -ln(104/ 724) / (5.87 * 10^-4)
Now, we can calculate the time:
t ≈ -ln(0.1437) / (5.87 * 10^-4)
Using a calculator or computer software, we find:
t ≈ 608.519 minutes
Therefore, it will take approximately 608.519 minutes for the partial pressure of cyclopropane to drop to below 104 torr.
To solve this problem, we need to use the first-order rate equation:
Rate = k * [cyclopropane]
Where:
- Rate is the rate of reaction
- k is the rate constant
- [cyclopropane] is the concentration of cyclopropane
We can rearrange the equation to solve for the concentration of cyclopropane:
[cyclopropane] = Rate / k
First, let's find the initial concentration of cyclopropane:
Pressure = 724 torr
Temperature = 485 degrees Celsius
We need to convert torr to atm and Celsius to Kelvin:
Pressure = 724 torr * (1 atm / 760 torr) = 0.95 atm
Temperature = 485 degrees Celsius + 273.15 = 758.15 K
Using the ideal gas law equation, we can find the concentration in molarity (mol/L):
PV = nRT
0.95 atm * 2.5 L = n * 0.0821 L·atm/(mol·K) * 758.15 K
n = (0.95 atm * 2.5 L) / (0.0821 L·atm/(mol·K) * 758.15 K) = 0.0375 mol
The concentration is then:
[cyclopropane] = 0.0375 mol / 2.5 L = 0.015 mol/L
Next, we need to find the rate of decrease of the concentration:
Rate = k * [cyclopropane] = 5.87*10^-4 min^-1 * 0.015 mol/L = 8.805*10^-6 mol/(L·min)
Now, we can find the time required for the partial pressure of cyclopropane to drop to below 104 torr:
Pressure = 104 torr
Pressure = 104 torr * (1 atm / 760 torr) = 0.137 atm
Using the ideal gas law, we can find the new concentration:
PV = nRT
0.137 atm * 2.5 L = n * 0.0821 L·atm/(mol·K) * 758.15 K
n = (0.137 atm * 2.5 L) / (0.0821 L·atm/(mol·K) * 758.15 K) = 0.00639 mol
The new concentration is then:
[cyclopropane] = 0.00639 mol / 2.5 L = 0.002556 mol/L
We can now calculate the time using the rate equation:
Rate = k * [cyclopropane]
t = [cyclopropane] / (k * Rate)
t = (0.015 mol/L - 0.002556 mol/L) / (5.87*10^-4 min^-1 * 8.805*10^-6 mol/(L·min))
t = 18.16 min
Therefore, it will take approximately 18.16 minutes for the partial pressure of cyclopropane to drop to below 104 torr.