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

Options

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

Explanation

To determine the enthalpy of the reaction

\[ C_2H_4(g) + 6F_2(g) \rightarrow 2CF_4(g) + 4HF(g) \]

we will use Hess's law and the given reactions with their enthalpy changes.

### 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 derive the target reaction.

- We need to form \( 2HF(g) \) from \( H_2(g) + F_2(g) \). Since we need \( 4HF(g) \) in the target reaction, we will multiply the first reaction by 2:

\[ 2H_2(g) + 2F_2(g) \rightarrow 4HF(g) \quad \Delta H = 2 \times (-537 \, \text{kJ}) = -1074 \, \text{kJ} \]

- The second reaction gives us \( CF_4(g) \). Since we need \( 2CF_4(g) \), we will multiply the second reaction by 2:

\[ 2C(s) + 4F_2(g) \rightarrow 2CF_4(g) \quad \Delta H = 2 \times (-680 \, \text{kJ}) = -1360 \, \text{kJ} \]

- The third reaction needs to be reversed to produce \( C(s) \) from \( C_2H_4(g) \):

\[ C_2H_4(g) \rightarrow 2C(s) + 2H_2(g) \quad \Delta H = -\Delta H_3^\circ = -52 \, \text{kJ} \]

### Step 3: Sum the reactions and their enthalpy changes.

Now we will add the modified reactions:

1. \( 2H_2(g) + 2F_2(g) \rightarrow 4HF(g) \) \(\Delta H = -1074 \, \text{kJ}\)

2. \( 2C(s) + 4F_2(g) \rightarrow 2CF_4(g) \) \(\Delta H = -1360 \, \text{kJ}\)

3. \( C_2H_4(g) \rightarrow 2C(s) + 2H_2(g) \) \(\Delta H = -52 \, \text{kJ}\)

Adding these reactions gives:

\[ C_2H_4(g) + 6F_2(g) \rightarrow 2CF_4(g) + 4HF(g) \]

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

Now we sum the enthalpy changes:

\[ \Delta H_{reaction} = (-1074) + (-1360) + (-52) \]

Calculating this:

\[ \Delta H_{reaction} = -1074 - 1360 - 52 = -2486 \, \text{kJ} \]

### Final Answer:

Thus, the enthalpy of the reaction at 298 K is

\[ \Delta H = -2486 \, \text{kJ} \]

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