Cyclopropane reacts to form propene in the gas phase. The reaction is first order in cylclopropane and has a rate constant of at 485 .
If a 2.5 reaction vessel initially contains 724 of cyclopropane at 485 , how long will it take for the partial pressure of cylclopropane to drop to below 104 ?
(if possible, can i be given the correct answer: i keep coming up with negatives)
You don't have units of k listed and you can't work the problem without them.. Show your work; perhaps we can find the error. That's better than us trying to figure out what you've done.
To solve this problem, we can use the first-order rate equation:
ln([A]₀/[A]) = kt
Where [A]₀ is the initial concentration of cyclopropane, [A] is the concentration of cyclopropane at a given time, k is the rate constant, and t is the time.
In this case, the concentration of cyclopropane is described in terms of partial pressure instead of molarity. However, since the reaction is taking place in the gas phase, we can use the ideal gas law to relate partial pressure and concentration:
PV = nRT
Where P is the partial pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
By rearranging the ideal gas law equation, we can express the concentration in terms of partial pressure:
[A] = P/(RT)
Now, we can substitute this expression for concentration into the rate equation and solve for time:
ln(([A]₀/P₀)/([A]/P)) = kt
ln(P₀/P) = kt
Solving for t:
t = ln(P₀/P)/(k)
Given:
Initial partial pressure, P₀ = 485 atm
Final partial pressure, P = 104 atm
Rate constant, k = 0.086 min⁻¹
Substituting these values:
t = ln(485/104)/(0.086)
Using a calculator, the natural logarithm of the ratio 485/104 is approximately 1.673.
t = 1.673/0.086
t ≈ 19.44 minutes
Therefore, it will take approximately 19.44 minutes for the partial pressure of cyclopropane to drop to below 104 atm in the given reaction vessel.