Use the given standard enthalpies of formation (in kJ/mol) to determine the enthalpy of reaction of the following reaction :
`NH_(3)(g)+3F_(2)(g)rarrNF_(3)+3HF(g)`
`DeltaH_(f)^(@)(NH_(3),g)=-46.2, " "DeltaH_(f)^(@)(NF_(3),g)=-113.0 ," "DeltaH_(f)^(@)(HF, g)=-269.0`

Options

A. `-335.8 kJ//mol`
B. `-873.8 kJ//mol`
C. `-697.2 kJ//mol`
D. `-890.4 kJ//mol`

Explanation

To determine the enthalpy of the reaction given by:

\[ \text{NH}_3(g) + 3\text{F}_2(g) \rightarrow \text{NF}_3(g) + 3\text{HF}(g) \]

we will use the standard enthalpies of formation provided for each substance involved in the reaction.

### Step-by-step solution:

1. **Identify the standard enthalpies of formation**:
- \(\Delta H_f^\circ(\text{NH}_3(g)) = -46.2 \, \text{kJ/mol}\)
- \(\Delta H_f^\circ(\text{NF}_3(g)) = -113.0 \, \text{kJ/mol}\)
- \(\Delta H_f^\circ(\text{HF}(g)) = -269.0 \, \text{kJ/mol}\)
- The standard enthalpy of formation for \(\text{F}_2(g)\) is \(0 \, \text{kJ/mol}\) since it is in its elemental form.

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

3. **Calculate the total enthalpy of formation for the products**:
- For \(\text{NF}_3\): \(1 \times (-113.0) = -113.0 \, \text{kJ}\)
- For \(3 \text{HF}\): \(3 \times (-269.0) = -807.0 \, \text{kJ}\)
- Total for products:
\[ \Delta H_f^\circ(\text{products}) = -113.0 + (-807.0) = -920.0 \, \text{kJ} \]

4. **Calculate the total enthalpy of formation for the reactants**:
- For \(\text{NH}_3\): \(1 \times (-46.2) = -46.2 \, \text{kJ}\)
- For \(3 \text{F}_2\): \(3 \times 0 = 0 \, \text{kJ}\)
- Total for reactants:
\[ \Delta H_f^\circ(\text{reactants}) = -46.2 + 0 = -46.2 \, \text{kJ} \]

5. **Substitute the values into the enthalpy change formula**:
\[ \Delta H_{reaction} = (-920.0) - (-46.2) = -920.0 + 46.2 = -873.8 \, \text{kJ} \]

### Final Answer:
The enthalpy of the reaction is:
\[ \Delta H_{reaction} = -873.8 \, \text{kJ} \]