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 find the activation energy (E_a) for the bimolecular recombination of two OH radicals to form H2O and O, we can follow these steps:

### Step 1: Understand the Reaction
We are given the reaction:
\[ H_2O + O \rightarrow 2OH \]
This reaction has a given enthalpy change (ΔH) of +72 kJ/mol and an activation energy (E_a) of 77 kJ/mol.

### Step 2: Identify the Reverse Reaction
The reverse reaction, which we need to analyze, is:
\[ 2OH \rightarrow H_2O + O \]
This is the bimolecular recombination of two OH radicals.

### Step 3: Use the Relationship Between E_a and ΔH
For a reaction, the relationship between the activation energy of the forward reaction (E_a forward), the activation energy of the backward reaction (E_a backward), and the enthalpy change (ΔH) is given by:
\[ \Delta H = E_a \text{(forward)} - E_a \text{(backward)} \]

### Step 4: Rearranging the Equation
We can rearrange the equation to find the activation energy for the backward reaction:
\[ E_a \text{(backward)} = E_a \text{(forward)} - \Delta H \]

### Step 5: Substitute the Known Values
Now we substitute the known values into the equation:
- E_a (forward) = 77 kJ/mol
- ΔH = 72 kJ/mol

So,
\[ E_a \text{(backward)} = 77 \text{ kJ/mol} - 72 \text{ kJ/mol} \]

### Step 6: Calculate E_a (backward)
Now we perform the calculation:
\[ E_a \text{(backward)} = 77 - 72 = 5 \text{ kJ/mol} \]

### Final Answer
The activation energy for the bimolecular recombination of two OH radicals to form H2O and O is:
\[ \boxed{5 \text{ kJ/mol}} \]

---