If the end energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ `mol^(-1)` respectively, then `DeltaH^(@)` for the reaction, `H_(2)(g)+Br_(2)(g)to2HB r(g)` is

Options

A. `-261 kJ`
B. `+103 kJ`
C. `+261 kJ`
D. `-103 kJ`

Answer

To calculate the enthalpy change (ΔH) for the reaction:

\[ H_2(g) + Br_2(g) \rightarrow 2 HBr(g) \]

we will use the bond energies provided:

- Bond energy of H-H = 433 kJ/mol
- Bond energy of Br-Br = 192 kJ/mol
- Bond energy of H-Br = 364 kJ/mol

### Step 1: Identify the bonds broken and formed
In the reaction, we break the following bonds:
- 1 H-H bond
- 1 Br-Br bond

And we form:
- 2 H-Br bonds (since 2 moles of HBr are produced)

### Step 2: Write the equation for ΔH
The enthalpy change for the reaction can be calculated using the formula:

\[ \Delta H = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]

### Step 3: Calculate the total bond energy of reactants
Total bond energy of reactants:
- H-H bond: 433 kJ/mol
- Br-Br bond: 192 kJ/mol

So, the total bond energy of reactants is:
\[ 433 + 192 = 625 \text{ kJ/mol} \]

### Step 4: Calculate the total bond energy of products
Total bond energy of products (for 2 moles of HBr):
- H-Br bond: 364 kJ/mol (for 1 HBr)
- Since we have 2 HBr, we multiply by 2:
\[ 2 \times 364 = 728 \text{ kJ/mol} \]

### Step 5: Substitute the values into the ΔH equation
Now we can substitute the values into the ΔH equation:
\[ \Delta H = 625 \text{ kJ/mol} - 728 \text{ kJ/mol} \]

### Step 6: Calculate ΔH
\[ \Delta H = 625 - 728 = -103 \text{ kJ/mol} \]

### Conclusion
The enthalpy change (ΔH) for the reaction \( H_2(g) + Br_2(g) \rightarrow 2 HBr(g) \) is:
\[ \Delta H = -103 \text{ kJ/mol} \]