Q3. Standard enthalpies of formation for atomic oxygen and for ozone are given h

Question

Q3. Standard enthalpies of formation for atomic oxygen and for ozone are given here:

O(g), fH° = 249.2 kJ mol-1.

O3(g), fH° = 142.7 kJ mol-1

For each of the reactions below, evaluate the standard reaction enthalpy (kJ mol-1) corresponding to the dissociation threshold, and therefore calculate the wavelength (nm) and state the “colour” of a photon with the minimum energy sufficient to achieve dissociation.

(a) dissociation of molecular oxygen, O2(g). {numerical ans. 240 nm}

(b) dissociation of ozone, O3(g), into O2(g) and O(g). {numerical ans. 1123 nm}

Explanation / Answer

SOLUTION:

a) O2(g) ----> 2O(g) DHrxn = 2*249.2 = +498.4 kj/mol

1 mole O2 = +498.4 kj

1 mole O2 = 6.023*10^23 molecules = 498.4 kj

1 molecule of O2 required energy = 498.4*10^3/(6.023*10^23)

                                     = 8.275*10^-19 joule

   E = hc/L

h = planks constant = 6.625*10^-34 j.s

c = light velocity = 3*10^8 m/s

L = wave length = ?

(8.275*10^-19) = (6.625*10^-34)*(3*10^8)/L

    L = wave length = 2.4*10^-7 m

                  = 240 nm ( no color, because it is not in vibgyor)

b)

         O3 ----> O2+O

    DHrxn = (249.2+0)-(142.7)

          = 106.5 kj/mol

1 mole O3 = 6.023*10^23 molecules = 106.5 kj

1 molecule of O3 required energy = 106.5 *10^3/(6.023*10^23)

                                 = 1.77*10^-19 joule

   E = hc/L

h = planks constant = 6.625*10^-34 j.s

c = light velocity = 3*10^8 m/s

L = wave length = ?

(1.77*10^-19) = (6.625*10^-34)*(3*10^8)/L

    L = wave length = 1.123*10^-6 m

                   = 1123 nm(no color , because it is not in vibgyor)

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