For the decomposition of gaseous dinitrogen pentoxide (shownbelow),
2 N2O5(g).....> 4NO2(g) + O2(g)
the rate constant is k = 2.8 10-3 s-1 at 60°C. The initial concentration of N2O5 is 1.52 mol/L.
(a) What is [N2O5] after 5.00 min?
_____mol/L
(b) What fraction of the N2O5 has decomposed after 5.00 min?
_____
1.01
0.57
The rate constant for the first order decomposition at 45 o
C of dinitrogen pentaoxide, N2O5, dissolved in
chloroform, CHCl3, is 6.2 x 10-4 min-1
2N2O5 → 4NO2 + O2
What is the rate of decomposition of N2O5 and rates of formation of NO2 and O2 when the concentration of N2O5 is
0.4 mol.dm-3
.
To find the concentration of N2O5 after 5.00 min, we can use the integrated rate law for a second-order reaction. In this case, the integrated rate law is:
[N2O5] = [N2O5]0 / (1 + kt)
Where [N2O5] is the concentration of N2O5 at a given time, [N2O5]0 is the initial concentration of N2O5, k is the rate constant, and t is the time in seconds.
(a) To find [N2O5] after 5.00 min, we need to convert the time to seconds. Since there are 60 seconds in a minute, 5.00 min is equal to 5.00 * 60 = 300 seconds.
Now we can plug in the values into the integrated rate law:
[N2O5] = 1.52 mol/L / (1 + (2.8 * 10^-3 s^-1 * 300 s))
Calculating this expression will give us the concentration of N2O5 after 5.00 min.
(b) To find the fraction of N2O5 that has decomposed after 5.00 min, we can subtract the concentration of N2O5 after 5.00 min from the initial concentration and divide by the initial concentration:
Fraction decomposed = ([N2O5]0 - [N2O5]) / [N2O5]0
Plugging in the values we found in part (a), we can calculate the fraction decomposed.
The unit in the problem for k is sec-1 which means this is a first order reaction.
ln(No/N) = kt
No = 1.52 M
N = ?
k = given
t = 5 min
Solve for N.
For part b, You know No and N, fraction remaining is N/No. Subtract from 1.0 to find fraction decomposed.
Post your work if you get stuck.