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1900's Physics and Wave

The reaction ...

Question

The reaction between $H_{2}$ and ${B r}_{2}$ to form HBr in the presence of the light is initiated by the photo decomposition of ${B r}_{2}$ into free Br atoms(free radicals) by absorption of light.The bond dissociation energy of ${B r}_{2}$ is 192 KJ/mole.What is the longest wavelength of the photon that would initiate the reaction?
(1 eV/atom $=$ 96KJ/mol; $h c =$ 1240 ev-nm)

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Solution

Bond dissociation enthalpy is the minimum amount of energy required to initiate bond dissociation.
$E = h c / λ$ where $λ$ is the wavelength of light required.
for constant hc; $λ$ is inversely proportional to E
i.e maximum wavelength for minimum energy $= h c / E = ((6.626 \times 10^{- 34} \times 3 \times 10^{8}) (6.02 \times 10^{23}) / 192 \times 10^{3} J = 6.23 \times 10^{- 7} m$

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H_{2}

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B r_{2}

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(1 \frac{e V}{a t o m} = 96 k j / m o l; h c = 1240 e v \times n m)

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e V \circ A

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(1 \frac{e v}{a t o m} 5 = 96 k J / m o l .)

λ_{1} = 350 n m

λ_{2} = 54 n m

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= \frac{1240}{λ (i n n m)} e V)

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