Question

The reaction between $H_2$ and $B{r}_2$ to form $HBr$ in presence of light is initiated by the photo decomposion of $B{r}_2$ into free $Br$ atoms (free redicals) by absorption of light. The bond dissociation energy of $B{r}_2$ is $Kj/mole$. What is the longest wavelength (in $\mathring{A}$) of the photon that would initiate the real $(1\frac{eV}{atom}=96kj/mol;hc=1240ev\times nm)$.

Answer 1

Bond dissociation enthalpy is the minimum amount of energy required to initiate bond dissociation.

$E=hc/\lambda$ where $\lambda$ is the wavelength of light required.

for constant hc; $\lambda$ is inversely proportional to E

i.e maximum wavelength for minimum energy$=hc/E=\left(\right(6.626\times {10}^{-34}\times 3\times {10}^8)\times 6.02\times {10}^{23})/192\times {10}^{3}J=6.2\times {10}^{-7}m$