Dinitrogen pentaoxide, N 2 O 5 (molar mass = 108.02 g/mol), decomposes by first-
ID: 573860 • Letter: D
Question
Dinitrogen pentaoxide, N2O5 (molar mass = 108.02 g/mol), decomposes by first-order kinetics with a half-life of 5.20 hours at 25ºC.
a) What is the rate constant for this decomposition at 25ºC?
b) If 5.50 grams of N2O5 is injected into a 2.00 Liter container at 25ºC, then how much will remain after 4.00 hours?
c) If the half-life of the reaction is 2.85 hours when the temperature is 35ºC, then what is the activation energy for the decomposition?
Explanation / Answer
(a)
Relation between rate constant and half life time of first order reaction is,
k = 0.693 / t1/2
k = 0.693 / 5.20
k = 0.133 h-1
(b)
Moles of N2O5 = mass / molar mass = 5.50 / 108.02 = 0.0509 mol
Molarity = moles of solute / volume of solution in L
M = 0.0509 / 2.00
M = 0.0254 M
Integrated rate constant eqaution for first order reaction kinetics is,
k = ( 1 / t ) * ln([A]0 / [A])
0.133 = ( 1 / 4.00 ) * ln(0.0254 / [A])
ln(0.0254 / [A]) = 0.532
0.0254 / [A] = e0.532
0.0254 / [A] = 1.70
[A] = remaining concentration after 4.00 hours = 0.0149 M
Moles of N2O5 = 0.0149 * 2.00 = 0.0298 mol
Mass of N2O5 remained after 4.00 hours = 0.0298 * 108.02 = 3.22 g.
(c)
k2 = 0.693 / 2.85
k2 = 0.243 h-1
Using Arrhenius equation,
lnk2/k1 = (Ea / R)*[(1/T1) - (1/T2)]
ln(0.243 / 0.133) = (Ea / 0.008314) * [(1/298.15) - (1/308.15)]
0.603 = Ea * 0.0131
Ea = activation energy = 46.1 kJ