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?

(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

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