# Why can UV light initiate a reaction between hydrogen and chlorine gas?

Can someone explain me how does UV light help combine chloride gas and hydrogen to produce hydrochloric acid?

$$\ce{Cl2(g) + H2(g) -> 2HCl(g)}$$

• This is a standard (photochemically induced) chain reaction involving the H and Cl radicals that occurs after the photon breaks the chlorine bond. Similar reactions to this are described in many phys. chem. textbooks where the standard (generic) scheme used describes the hydrogen/bromine reaction. – porphyrin Feb 19 '17 at 12:07

$\ce{Cl-Cl}$ bond in $\ce{Cl2}$ is weak enough to be broken by mere UV rays (present in sunlight), and hence they undergo homolytic cleavage (the resultant products are $\ce{Cl}$ atoms, not ions) to form two $\ce{Cl}$ free radicals. Now these $\ce{Cl}$ free radicals are extremely reactive (due to one free electron that it can share in a covalent bond) and hence attack hydrogen gas.
Now is the time you consider energy lowering. $\ce{H-Cl}$ bond is lot more stronger than $\ce{H2}$, and hence the formation of $\ce{H-Cl}$ bond formation will be favored instead of $\ce{H-H}$. Therefore, $\ce{Cl}$ free radical attacks $\ce{H2}$ gas to break the bond and form $\ce{HCl}$. The other $\ce{H}$ atom hence liberated forms a bond with the other $\ce{Cl}$ free radical.
• The HCl bond is actually slightly weaker ( $35728 \pu{cm^{-1}}$ ) than the $\ce{H2}$ bond ( $36099 \pu{cm^{-1}}$ ) . The H atoms are formed with chain $\ce{Cl + H2 <=> HCl + H}$. The H reacts rapidly in chain steps $\ce{H + Cl2 <=> HCl + Cl }$ and $\ce{H + HCl <=> H2 + Cl}$ so its concentration is always low. The termination step is $\ce{Cl + Cl +M <=> Cl2 +M}$, M being any un-reactive gas. [H] is too low for its equivalent termination step to be important. The photochemical chain reaction can be explosive and makes a good demonstration. – porphyrin Feb 19 '17 at 15:12