There is formation of `H_(2)O_(2)` in the upper atomsphere.
`H_(2)O + O rarr 2OH rarr H_(2)O_(2)`
`Delta H = 72 kJ mol^(-1) , E_(a) = 77 kJ mol^(-1)`
What will be the `E_(a)` for bimolecular recombination of two OH radicals to form `H_(2)O` and O in kJ.

Answer

To solve the problem, we need to find the activation energy (E_a) for the bimolecular recombination of two OH radicals to form H₂O and O. We are given the following information:

- ΔH (enthalpy change) = 72 kJ/mol
- E_a (activation energy for the forward reaction) = 77 kJ/mol

We can use the relationship between the activation energies of the forward and backward reactions and the enthalpy change:

1. **Understand the relationship**:
The relationship between the activation energies and the enthalpy change is given by:
\[ E_a(\text{backward}) = E_a(\text{forward}) - \Delta H \]
Here, E_a(backward) is the activation energy for the reaction going from products to reactants (in this case, from H₂O and O to 2 OH).

2. **Substitute the known values**:
We can substitute the known values into the equation:
\[ E_a(\text{backward}) = 77 \, \text{kJ/mol} - 72 \, \text{kJ/mol} \]

3. **Calculate E_a(backward)**:
\[ E_a(\text{backward}) = 77 - 72 = 5 \, \text{kJ/mol} \]

Thus, the activation energy for the bimolecular recombination of two OH radicals to form H₂O and O is **5 kJ/mol**.