Given the following equations and `DeltaH^(@)` values, determine the enthalpy of reaction at 298 K for the reaction :
`C_(2)H_(4)(g)+6F_(2)(g) to 2HF_(4)(g)+4HF(g)`
`H_(2)(g)+F_(2)(g)to2HF(g)+2HF(g)," "DeltaH_(1)^(@)=-537" kJ"`
`C(s)+2F_(2)(g)toCF_(4)(g)," "DeltaH_(2)^(@)=-680" kJ"`
`2C(s)+2H_(2)(g)toC_(2)H_(4)(g)," "DeltaH_(3)^(@)=52" kJ"`

Options

A. `-1165`
B. `-2486`
C. `+1165`
D. `+2486`

Explanation

To determine the enthalpy of the reaction \( C_2H_4(g) + 6F_2(g) \rightarrow 2CF_4(g) + 4HF(g) \) using the given reactions and their enthalpy changes, we will follow these steps:

### Step 1: Write down the given reactions and their enthalpy changes.

1. \( H_2(g) + F_2(g) \rightarrow 2HF(g) \), \( \Delta H_1^\circ = -537 \, \text{kJ} \)
2. \( C(s) + 2F_2(g) \rightarrow CF_4(g) \), \( \Delta H_2^\circ = -680 \, \text{kJ} \)
3. \( 2C(s) + 2H_2(g) \rightarrow C_2H_4(g) \), \( \Delta H_3^\circ = 52 \, \text{kJ} \)

### Step 2: Manipulate the reactions to match the target reaction.

We need to manipulate the given reactions to derive the target reaction.

1. **For Reaction 1**: We need 4 moles of \( HF \). Therefore, we will multiply the entire reaction by 2:
\[ 2H_2(g) + 2F_2(g) \rightarrow 4HF(g), \quad \Delta H = 2 \times (-537) = -1074 \, \text{kJ} \]

2. **For Reaction 2**: We need 2 moles of \( CF_4 \). Therefore, we will multiply the entire reaction by 2:
\[ 2C(s) + 4F_2(g) \rightarrow 2CF_4(g), \quad \Delta H = 2 \times (-680) = -1360 \, \text{kJ} \]

3. **For Reaction 3**: We need to reverse this reaction because \( C_2H_4 \) is a reactant in our target reaction:
\[ C_2H_4(g) \rightarrow 2C(s) + 2H_2(g), \quad \Delta H = -52 \, \text{kJ} \]

### Step 3: Add the manipulated reactions.

Now we will add the manipulated reactions together:

1. From Reaction 1:
\[ 2H_2(g) + 2F_2(g) \rightarrow 4HF(g) \quad \Delta H = -1074 \, \text{kJ} \]

2. From Reaction 2:
\[ 2C(s) + 4F_2(g) \rightarrow 2CF_4(g) \quad \Delta H = -1360 \, \text{kJ} \]

3. From Reaction 3 (reversed):
\[ C_2H_4(g) \rightarrow 2C(s) + 2H_2(g) \quad \Delta H = +52 \, \text{kJ} \]

Adding these reactions together:
\[ C_2H_4(g) + 6F_2(g) \rightarrow 2CF_4(g) + 4HF(g) \]

### Step 4: Calculate the total enthalpy change.

Now we sum the enthalpy changes:
\[ \Delta H = -1074 + (-1360) + 52 = -2382 \, \text{kJ} \]

### Final Answer:
The enthalpy of the reaction at 298 K is:
\[ \Delta H = -2382 \, \text{kJ} \]