Chloroform is passed from methane according to the reaction `:`
`CH_(4)(g) + 3Cl_(2)(g) rarr CHCl_(3)(l) + 3HCl(g)`
Calculate `Delta H ` for the reaction given that enthalpies of formation of HCl(g), `CH_(4)(g) ` and `CHCl_(3)(g)` are - 92.0 ,-74.9 and -134.3kJ per mole respectively.

Explanation

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

\[ \text{CH}_4(g) + 3\text{Cl}_2(g) \rightarrow \text{CHCl}_3(l) + 3\text{HCl}(g) \]

we will use the enthalpies of formation (ΔHf) of the reactants and products.

### Step-by-step Solution:

1. **Identify the Enthalpies of Formation:**
- ΔHf for CH₄(g) = -74.9 kJ/mol
- ΔHf for HCl(g) = -92.0 kJ/mol
- ΔHf for CHCl₃(l) = -134.3 kJ/mol
- ΔHf for Cl₂(g) = 0 kJ/mol (as it is in its standard state)

2. **Write the Formula for ΔH:**
The change in enthalpy for the reaction can be calculated using the formula:
\[ \Delta H = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

3. **Calculate the Enthalpy of Products:**
- For CHCl₃(l): ΔHf = -134.3 kJ/mol
- For 3 HCl(g): 3 × ΔHf = 3 × (-92.0 kJ/mol) = -276.0 kJ/mol
- Total for products:
\[ \Delta H_f \text{ (products)} = -134.3 + (-276.0) = -410.3 \text{ kJ/mol} \]

4. **Calculate the Enthalpy of Reactants:**
- For CH₄(g): ΔHf = -74.9 kJ/mol
- For 3 Cl₂(g): 3 × ΔHf = 3 × 0 = 0 kJ/mol
- Total for reactants:
\[ \Delta H_f \text{ (reactants)} = -74.9 + 0 = -74.9 \text{ kJ/mol} \]

5. **Calculate ΔH for the Reaction:**
Now, substituting the values into the ΔH formula:
\[ \Delta H = \Delta H_f \text{ (products)} - \Delta H_f \text{ (reactants)} \]
\[ \Delta H = -410.3 - (-74.9) = -410.3 + 74.9 = -335.4 \text{ kJ/mol} \]

### Final Result:
\[ \Delta H = -335.4 \text{ kJ/mol} \]