Using the equations
2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol
Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol

Determine the molar enthalpy (in kJ/mol) for the reaction
3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s)

To determine the molar enthalpy of the reaction:

\[
3 \text{SiCl}_4 (s) + 4 \text{Fe} (s) \rightarrow 4 \text{FeCl}_3 (s) + 3 \text{Si} (s)
\]

we can use Hess's Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of individual steps that lead to the final reaction.

### Given Equations:
1. \( 2 \text{Fe} (s) + 3 \text{Cl}_2 (g) \rightarrow 2 \text{FeCl}_3 (s) \)  \( \Delta H^\circ = -800.0 \, \text{kJ/mol} \)
2. \( \text{Si} (s) + 2 \text{Cl}_2 (g) \rightarrow \text{SiCl}_4 (s) \) \( \Delta H^\circ = -640.1 \, \text{kJ/mol} \)

### Step-by-Step Solution:

#### Step 1: Reverse the second reaction
We need to reverse the second reaction because in the desired reaction, SiCl₄ is being consumed, not produced.

\[
\text{SiCl}_4 (s) \rightarrow \text{Si} (s) + 2 \text{Cl}_2 (g)
\]
When we reverse the reaction, the enthalpy change will also reverse its sign:
\[
\Delta H^\circ = +640.1 \, \text{kJ/mol}
\]

#### Step 2: Multiply the reactions to match the desired equation
We need to multiply the reactions by the appropriate coefficients to match the stoichiometry of the target reaction.

1. The first reaction (for Fe and FeCl₃) needs to be multiplied by 2, so that we get 4 Fe and 4 FeCl₃:
\[
4 \text{Fe} (s) + 6 \text{Cl}_2 (g) \rightarrow 4 \text{FeCl}_3 (s)
\]
\[
\Delta H^\circ = 2 \times (-800.0 \, \text{kJ/mol}) = -1600.0 \, \text{kJ/mol}
\]

2. The second reversed reaction (for SiCl₄ and Si) needs to be multiplied by 3, so that we get 3 SiCl₄ and 3 Si:
\[
3 \text{SiCl}_4 (s) \rightarrow 3 \text{Si} (s) + 6 \text{Cl}_2 (g)
\]
\[
\Delta H^\circ = 3 \times (+640.1 \, \text{kJ/mol}) = +1920.3 \, \text{kJ/mol}
\]

#### Step 3: Add the reactions
Now, we can add the two reactions:

\[
3 \text{SiCl}_4 (s) + 4 \text{Fe} (s) \rightarrow 4 \text{FeCl}_3 (s) + 3 \text{Si} (s)
\]

The enthalpy change for the overall reaction is the sum of the enthalpy changes of the two reactions:

\[
\Delta H^\circ = -1600.0 \, \text{kJ/mol} + 1920.3 \, \text{kJ/mol} = +320.3 \, \text{kJ/mol}
\]

Thus, the molar enthalpy of the reaction is \( +320.3 \, \text{kJ/mol} \).

Would you like more details or clarification on any steps?

### 5 Related Questions:
1. How does Hess's Law apply to reactions involving multiple steps?
2. What is the significance of reversing reactions in thermodynamic calculations?
3. Can we always add and subtract enthalpy values in chemical reactions?
4. What is the difference between enthalpy of formation and reaction enthalpy?
5. How would the calculation change if different coefficients were used?

### Tip:
Always remember that when you reverse a reaction, you must also reverse the sign of the enthalpy change.

Using the equations 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol, Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol, determine the molar enthalpy (in kJ/mol) for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s).

Learn how to calculate the molar enthalpy for the reaction 3 SiCl₄ + 4 Fe → 4 FeCl₃ + 3 Si using Hess's Law. This detailed explanation covers key concepts such as reversing reactions and using enthalpy values, with step-by-step guidance for solving this thermodynamics problem.

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