What would be the concentration of dinitrogen pentoxide after 6.00 X 10(2) seconds and how long would it take for the concentration of N2O2 to decrease to 10.0% of its initial value in the following equation...
N2O5 to NO2 and O2
initial concentration is 1.65 X 10-2mol/L
rate constant is 4.80 X 10-4/s
time is 825s
According to my book, the answer is 1.24 X 10-2 mol/L and t= 4.80 X 10 (3)s
I have no idea how to set this problem up with the 1st order rate law...
This is a first order reaction (you know that from the units given for k).
ln(No/N) = kt
ln(0.0165/N) = 4.80E-4*600
Solve for N.
Use the same equation. No is the same, N is 10% of that, k is the same, solve for t.
Check my work for typos.
A vessel containing 39.5 cm3 of helium gas at 25°C and 106 kPa was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required 17.8 cm3 of ethanol, what was the final temperature of the helium?
many mols He were in the container? That's PV = nRT. Solve for n. Use that n in another PV = nRT. You know P,
V = 39.5cc - 17.8cc
R and solve for T.
Rememberr V must be i L, R is 0.08206 if you use P in atm or 8.314 if you use P in kPa. T must be in kelvin.
You should not piggy back your question on another post. Go to the top of the page and click on "Post a New Question."
So, to solve for t would I set up the problem like this...
ln (0.00124/0.0165)=4.80 X 10-4 X t?
To solve this problem using the first-order rate law, you can start by writing the rate equation for the reaction:
Rate = k[N2O5]
Where:
- Rate is the rate of the reaction (change in concentration per unit time),
- k is the rate constant of the reaction,
- [N2O5] is the concentration of dinitrogen pentoxide.
First, let's calculate the concentration of N2O5 after 6.00 X 10^2 seconds:
t = 6.00 X 10^2 seconds
initial concentration ([N2O5]0) = 1.65 X 10^-2 mol/L
rate constant (k) = 4.80 X 10^-4/s
We can use the integrated form of the first-order rate law equation to solve for the concentration at a given time:
[N2O5] = [N2O5]0 * exp(-kt)
Where:
- exp is the exponential function,
- k is the rate constant,
- t is the time.
Substituting the given values into the equation:
[N2O5] = (1.65 X 10^-2 mol/L) * exp(-(4.80 X 10^-4/s) * (6.00 X 10^2 s))
Calculating this expression will give you the concentration of N2O5 after 6.00 X 10^2 seconds.
Next, let's calculate the time it takes for the concentration of N2O5 to decrease to 10.0% of its initial value. In other words, we need to find the time (t) when [N2O5] = 0.1 * [N2O5]0 (where [N2O5]0 is the initial concentration).
Using the same integrated form of the first-order rate law equation:
[N2O5] = [N2O5]0 * exp(-kt)
Setting [N2O5] = 0.1 * [N2O5]0:
0.1 * [N2O5]0 = [N2O5]0 * exp(-kt)
Divide both sides by [N2O5]0:
0.1 = exp(-kt)
Taking the natural logarithm (ln) of both sides:
ln(0.1) = -kt
Rearranging the equation to solve for time (t):
t = -(ln(0.1)) / k
Substituting the given rate constant (k = 4.80 X 10^-4/s) into the equation will give you the time it takes for the concentration of N2O5 to decrease to 10.0% of its initial value.
By following these steps, you should be able to obtain the answer provided in your book: [N2O5] = 1.24 X 10^-2 mol/L and t = 4.80 X 10^3 s.