Given, `H_(2)(g)+Br_(2)(g)rarr2HBr(g),DeltaH_(1)^(0)` and standard enthalpy of condensaion of bromine is `DeltaH_(2)^(0)`, standard enthalpy of formation of `HBr` at `25^(@)C` is

Options

A. `DeltaH_(1)^(0)//2`
B. `DeltaH_(1)^(0)//2+DeltaH_(2)^(0)`
C. `DeltaH_(1)^(0)//2-DeltaH_(2)^(0)`
D. `(DeltaH_(1)^(0)-DeltaH_(2)^(0))//2`

Explanation

To find the standard enthalpy of formation of HBr at 25°C, we can use the given information about the reaction and the enthalpy values. Here is the step-by-step solution:

### Step 1: Write the Reaction
The reaction given is:
\[ H_2(g) + Br_2(g) \rightarrow 2 HBr(g) \]
This reaction has an associated enthalpy change, denoted as \( \Delta H_1^{0} \).

### Step 2: Understand Enthalpy of Formation
The standard enthalpy of formation of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states. For HBr, the formation reaction can be represented as:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Br_2(g) \rightarrow HBr(g) \]
The enthalpy change for this reaction is what we are trying to find, denoted as \( \Delta H_f^{0} \).

### Step 3: Relate the Enthalpy Changes
From the reaction, we can see that the formation of 2 moles of HBr corresponds to the enthalpy change \( \Delta H_1^{0} \). Therefore, the enthalpy change for the formation of 1 mole of HBr will be half of \( \Delta H_1^{0} \):
\[ \Delta H_f^{0} = \frac{1}{2} \Delta H_1^{0} \]

### Step 4: Include the Enthalpy of Condensation
The standard enthalpy of condensation of bromine, denoted as \( \Delta H_2^{0} \), is the energy change when bromine gas condenses into liquid bromine. Since we are dealing with gaseous reactants and products, we need to account for this condensation in our calculation.

The enthalpy of formation of HBr can be adjusted by subtracting the enthalpy of condensation of bromine:
\[ \Delta H_f^{0} = \frac{1}{2} \Delta H_1^{0} - \Delta H_2^{0} \]

### Step 5: Final Expression
Thus, the final expression for the standard enthalpy of formation of HBr at 25°C is:
\[ \Delta H_f^{0} = \frac{1}{2} \Delta H_1^{0} - \Delta H_2^{0} \]

### Conclusion
The standard enthalpy of formation of HBr at 25°C is given by:
\[ \Delta H_f^{0} = \frac{1}{2} \Delta H_1^{0} - \Delta H_2^{0} \]