Question

using the equations
\ce{h2(g) + f2(g) -> 2 hf(g)} \quad \delta h^\circ = -79.2\\,\text{kj/mol}
\ce{c(s) + 2 f2(g) -> cf4(g)} \quad \delta h^\circ = +141.3\\,\text{kj/mol}
\ce{2 c(s) + 2 h2(g) -> c2h4(g)} \quad \delta h^\circ = 52.4\\,\text{kj/mol}
determine the molar enthalpy (in kj/mol) for the reaction
\ce{c2h4(g) + 6 f2(g) -> 2 cf4(g) + 4 hf(g)}

Explanation:

Step1: Label the given reactions

Let's call the reactions:
1. \( \ce{H2(g) + F2(g) -> 2HF(g)} \quad \Delta H^\circ_1 = -79.2\ \text{kJ/mol} \)
2. \( \ce{C(s) + 2F2(g) -> CF4(g)} \quad \Delta H^\circ_2 = +141.3\ \text{kJ/mol} \)
3. \( \ce{2C(s) + 2H2(g) -> C2H4(g)} \quad \Delta H^\circ_3 = 52.4\ \text{kJ/mol} \)
And the target reaction: \( \ce{C2H4(g) + 6F2(g) -> 2CF4(g) + 4HF(g)} \)

Step2: Manipulate reaction 1

We need 4 moles of \( \ce{HF} \), so multiply reaction 1 by 2:
\( 2\times(\ce{H2(g) + F2(g) -> 2HF(g)}) \)
New reaction: \( \ce{2H2(g) + 2F2(g) -> 4HF(g)} \)
\( \Delta H^\circ_{1\times2} = 2\times(-79.2) = -158.4\ \text{kJ/mol} \)

Step3: Manipulate reaction 2

We need 2 moles of \( \ce{CF4} \), so multiply reaction 2 by 2:
\( 2\times(\ce{C(s) + 2F2(g) -> CF4(g)}) \)
New reaction: \( \ce{2C(s) + 4F2(g) -> 2CF4(g)} \)
\( \Delta H^\circ_{2\times2} = 2\times(141.3) = +282.6\ \text{kJ/mol} \)

Step4: Manipulate reaction 3

We need to reverse reaction 3 to get \( \ce{C2H4(g)} \) on the reactant side:
\( \ce{C2H4(g) -> 2C(s) + 2H2(g)} \)
\( \Delta H^\circ_{3\text{ reversed}} = -52.4\ \text{kJ/mol} \)

Step5: Add the manipulated reactions

Now add the three manipulated reactions:
- \( \ce{2H2(g) + 2F2(g) -> 4HF(g)} \quad \Delta H = -158.4 \)
- \( \ce{2C(s) + 4F2(g) -> 2CF4(g)} \quad \Delta H = +282.6 \)
- \( \ce{C2H4(g) -> 2C(s) + 2H2(g)} \quad \Delta H = -52.4 \)

Adding them together:
\( \ce{C2H4(g) + 2F2(g) + 4F2(g) + 2H2(g) + 2C(s) -> 4HF(g) + 2CF4(g) + 2C(s) + 2H2(g)} \)
Simplify (cancel out \( 2\ce{H2(g)} \) and \( 2\ce{C(s)} \)):
\( \ce{C2H4(g) + 6F2(g) -> 2CF4(g) + 4HF(g)} \) (which is the target reaction)

Now calculate the total \( \Delta H \):
\( \Delta H = -158.4 + 282.6 - 52.4 \)
First, \( -158.4 + 282.6 = 124.2 \)
Then, \( 124.2 - 52.4 = 71.8 \) Wait, no, wait: Wait, let's recalculate:

Wait, \( -158.4 + 282.6 = 124.2 \); then \( 124.2 - 52.4 = 71.8 \)? Wait, no, that can't be right. Wait, let's check the signs again.

Wait, reaction 1 multiplied by 2: \( \Delta H = 2*(-79.2) = -158.4 \)

Reaction 2 multiplied by 2: \( \Delta H = 2*(141.3) = 282.6 \)

Reaction 3 reversed: \( \Delta H = -52.4 \)

Now sum them: \( -158.4 + 282.6 - 52.4 \)

Let's compute step by step:

\( -158.4 - 52.4 = -210.8 \)

Then \( -210.8 + 282.6 = 71.8 \)? Wait, no, that's not correct. Wait, maybe I made a mistake in the manipulation.

Wait, target reaction: \( \ce{C2H4(g) + 6F2(g) -> 2CF4(g) + 4HF(g)} \)

Let's list the reactants and products:

Reactants: \( \ce{C2H4} \), \( 6\ce{F2} \)

Products: \( 2\ce{CF4} \), \( 4\ce{HF} \)

From reaction 3: \( \ce{2C + 2H2 -> C2H4} \), so reverse it: \( \ce{C2H4 -> 2C + 2H2} \) (correct, \( \Delta H = -52.4 \))

From reaction 2: \( \ce{C + 2F2 -> CF4} \), so 2 moles: \( \ce{2C + 4F2 -> 2CF4} \) (correct, \( \Delta H = 2*141.3 = 282.6 \))

From reaction 1: \( \ce{H2 + F2 -> 2HF} \), so 2 moles of \( \ce{H2} \) (from reversed reaction 3) would react with 2 moles of \( \ce{F2} \) to make 4 moles of \( \ce{HF} \): \( \ce{2H2 + 2F2 -> 4HF} \) (correct, \( \Delta H = 2*(-79.2) = -158.4 \))

Now, the \( \ce{F2} \) from reaction 2 (4 moles) and reaction 1 (2 moles) gives 6 moles, which matches the target. The \( \ce{C2H4} \) is from reversed reaction 3, the \( \ce{2C} \) from reversed reaction 3 cancels with \( \ce{2C} \) from reaction 2, the \( \ce{2H2} \) from reversed reaction 3 cancels with \( \ce{2H2} \) from reaction 1. So the sum is correct.

Now calculate \( \Delta H \):

\( -52.4 + 282.6 - 158.4 \)

First, \( 282.6 - 52.4 = 230.2 \)

Then, \( 230.2 - 158.4 = 71.8 \)? Wait, that seems low. Wait, maybe I messed up the signs. Wait, reaction 1 is exothermic (negative), reaction 2 is endothermic (positive), reaction 3 is endothermic (positive), so reversing it makes it exothermic (negative).

Wait, let's check the target reaction's enthalpy. Let's do the calculation again:

\( \Delta H = (2\times\Delta H_2) + (2\times\Delta H_1) - \Delta H_3 \)

Because:

- To get 2 \( \ce{CF4} \): 2*reaction 2 (so 2*ΔH2)
- To get 4 \( \ce{HF} \): 2*reaction 1 (so 2*ΔH1)
- To get \( \ce{C2H4} \) on the left: reverse reaction 3 (so -ΔH3)

So:

\( \Delta H = 2*(141.3) + 2*(-79.2) - 52.4 \)

Calculate each term:

2*141.3 = 282.6

2*(-79.2) = -158.4

-52.4

Now sum: 282.6 - 158.4 - 52.4 = (282.6) - (158.4 + 52.4) = 282.6 - 210.8 = 71.8? Wait, that's what I got. But let's check with another approach.

Alternatively, use Hess's law: the sum of the enthalpies of the manipulated reactions is the enthalpy of the target reaction.

Reaction 1 (x2): \( 2\ce{H2} + 2\ce{F2} -> 4\ce{HF} \), ΔH = -158.4

Reaction 2 (x2): \( 2\ce{C} + 4\ce{F2} -> 2\ce{CF4} \), ΔH = +282.6

Reaction 3 (reversed): \( \ce{C2H4} -> 2\ce{C} + 2\ce{H2} \), ΔH = -52.4

Now add all three:

\( 2\ce{H2} + 2\ce{F2} + 2\ce{C} + 4\ce{F2} + \ce{C2H4} -> 4\ce{HF} + 2\ce{CF4} + 2\ce{C} + 2\ce{H2} \)

Cancel out 2\ce{H2} and 2\ce{C} from both sides:

\( \ce{C2H4} + 6\ce{F2} -> 2\ce{CF4} + 4\ce{HF} \) (target reaction)

Now sum the ΔH: -158.4 + 282.6 - 52.4 = 71.8 kJ/mol. Wait, but let me check with actual values.

Wait, maybe I made a mistake in the multiplication. Let's re-express the target reaction in terms of the given reactions.

Target reaction: \( \ce{C2H4 + 6F2 -> 2CF4 + 4HF} \)

Let's express each compound:

- \( \ce{C2H4} \): comes from reaction 3 reversed: \( \ce{C2H4 -> 2C + 2H2} \) (so -ΔH3)
- \( \ce{CF4} \): comes from reaction 2 multiplied by 2: \( 2\ce{C + 4F2 -> 2CF4} \) (so +2ΔH2)
- \( \ce{HF} \): comes from reaction 1 multiplied by 2: \( 2\ce{H2 + 2F2 -> 4HF} \) (so +2ΔH1)
- \( \ce{F2} \): total F2 is 4 (from reaction 2x2) + 2 (from reaction 1x2) = 6, which matches.
- \( \ce{H2} \): 2 from reversed reaction 3, 2 from reaction 1x2: cancels.
- \( \ce{C} \): 2 from reversed reaction 3, 2 from reaction 2x2: cancels.

So the enthalpy is:

ΔH = (-ΔH3) + (2ΔH2) + (2ΔH1)

Plug in the numbers:

ΔH = (-52.4) + (2*141.3) + (2*(-79.2))

Calculate:

2*141.3 = 282.6

2*(-79.2) = -158.4

So:

ΔH = -52.4 + 282.6 - 158.4 = (282.6 - 158.4) - 52.4 = 124.2 - 52.4 = 71.8 kJ/mol.

Wait, but let me check with another method. Let's compute the enthalpy of formation for each compound, but since we don't have formation enthalpies, Hess's law is the way. So I think the calculation is correct.

Answer:

\( \boxed{71.8} \) (in kJ/mol)