Chemistry, asked by Karthick238, 1 year ago

The dissociation energy of h2 is 482.5kj/mol if h2 is exposed to radiant energy of wavelength 124nm , what % of radiant energy

Answers

Answered by AneesKakar
6

Answer:

87.15 %.

Explanation:

We know that the bond dissociation energy, which is known as the H-H bond energy which is required for breaking the H2 molecule into hydrogen atoms.

We know the formulae of radiant energy which is :-

hc/λ=6.626×10^−34×3×10^8/124×10^−9 m= 1.6×10^−18.

Hence, radiant energy per mol will be 1.6×10^−18 x 6.022 x 10^23 mol-1 =965365.4 J mol-1 .

In kilo joule we will get radiant energy per mol​ = 965.36 KJ mol-1.

Again, we know that K.E. is the Radiant energy - Dissociation energy .

     =  965.36 - 124 .

     = 841.36 KJ mol-1 .

So, the percentage of the radiant energy converted will be  K.E. is = 841.36/965.36×100 which on calculating we will get the value of radiant energy as 87.15 %.

                                         

Answered by Anonymous
2

Answer:

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Explanation:

We know that the bond dissociation energy, which is known as the H-H bond energy which is required for breaking the H2 molecule into hydrogen atoms.

We know the formulae of radiant energy which is :-

hc/λ=6.626×10^−34×3×10^8/124×10^−9 m= 1.6×10^−18.

Hence, radiant energy per mol will be 1.6×10^−18 x 6.022 x 10^23 mol-1 =965365.4 J mol-1

.

In kilo joule we will get radiant energy per mol = 965.36 KJ mol-1.

Again, we know that K.E. is the Radiant energy - Dissociation energy

.

     =  965.36 - 124

.

     = 841.36 KJ mol-1

.

So, the percentage of the radiant energy converted will be  K.E. is = 841.36/965.36×100 which on calculating we will get the value of radiant energy as 87.15 %.

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