Question 10 (10.97 points) 
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:

P4(g) + 10 Cl2(g) → 4 PCl5(s)	ΔH°rxn = ?
Given:
PCl5(s) → PCl3(g) + Cl2(g)	ΔH°rxn = +157 kJ
P4(g) + 6 Cl2(g) → 4 PCl3(g)	ΔH°rxn = -1207 kJ
Question 10 options:

-1786 kJ

-2100. kJ

-1835 kJ

-1364 kJ

-1050. kJ


To solve for the standard enthalpy of reaction, \( \Delta H^\circ_{\text{rxn}} \), for the reaction:

\[
\text{P}_4(g) + 10 \, \text{Cl}_2(g) \rightarrow 4 \, \text{PCl}_5(s)
\]

we can use the given reactions and manipulate them to match the target reaction. Here's the approach:

### Step 1: Write the Given Reactions with Their Enthalpies

1. **Reaction 1:**  
   \[
   \text{PCl}_5(s) \rightarrow \text{PCl}_3(g) + \text{Cl}_2(g) \quad \Delta H^\circ_{\text{rxn}} = +157 \, \text{kJ}
   \]

2. **Reaction 2:**  
   \[
   \text{P}_4(g) + 6 \, \text{Cl}_2(g) \rightarrow 4 \, \text{PCl}_3(g) \quad \Delta H^\circ_{\text{rxn}} = -1207 \, \text{kJ}
   \]

### Step 2: Manipulate the Reactions

We want to end up with \( \text{P}_4(g) + 10 \, \text{Cl}_2(g) \rightarrow 4 \, \text{PCl}_5(s) \).

To achieve this:

1. Reverse Reaction 1 to produce \( 4 \, \text{PCl}_5(s) \) as a product. Reversing the reaction changes the sign of \( \Delta H^\circ_{\text{rxn}} \):
   \[
   4 \, \text{PCl}_3(g) + 4 \, \text{Cl}_2(g) \rightarrow 4 \, \text{PCl}_5(s) \quad \Delta H^\circ_{\text{rxn}} = -4 \times 157 = -628 \, \text{kJ}
   \]

2. Use Reaction 2 as given:
   \[
   \text{P}_4(g) + 6 \, \text{Cl}_2(g) \rightarrow 4 \, \text{PCl}_3(g) \quad \Delta H^\circ_{\text{rxn}} = -1207 \, \text{kJ}
   \]

### Step 3: Add the Reactions

Now, add the manipulated reactions:

\[
\begin{align*}
\text{P}_4(g) + 6 \, \text{Cl}_2(g) &\rightarrow 4 \, \text{PCl}_3(g) \quad (\Delta H^\circ = -1207 \, \text{kJ}) \\
4 \, \text{PCl}_3(g) + 4 \, \text{Cl}_2(g) &\rightarrow 4 \, \text{PCl}_5(s) \quad (\Delta H^\circ = -628 \, \text{kJ})
\end{align*}
\]

### Step 4: Combine and Simplify

Adding these reactions gives:

\[
\text{P}_4(g) + 10 \, \text{Cl}_2(g) \rightarrow 4 \, \text{PCl}_5(s)
\]

And the enthalpy change for the overall reaction is:

\[
\Delta H^\circ_{\text{rxn}} = -1207 \, \text{kJ} + (-628 \, \text{kJ}) = -1835 \, \text{kJ}
\]

### Answer

The enthalpy change \( \Delta H^\circ_{\text{rxn}} \) for the reaction is:

\[
\boxed{-1835 \, \text{kJ}}
\]

### Correct Option:

-1835 kJ

---

Would you like a further breakdown of each manipulation step? Here are some related questions:

1. How do we determine whether to reverse a reaction when finding enthalpies?
2. What is the significance of a negative \( \Delta H^\circ_{\text{rxn}} \)?
3. Why does adding reactions' enthalpies work in this solution?
4. How does Hess's Law help in calculating reaction enthalpies?
5. What are common mistakes to avoid when combining enthalpies?

**Tip:** Always check if reversing a reaction or multiplying it by a coefficient is necessary to match the desired final reaction.

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:

P4(g) + 10 Cl2(g) → 4 PCl5(s) ΔH°rxn = ?

Given:
PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = +157 kJ
P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ

Options:
-1786 kJ
-2100 kJ
-1835 kJ
-1364 kJ
-1050 kJ

Learn how to determine the standard enthalpy change (ΔH°rxn) for the reaction P4(g) + 10 Cl2(g) → 4 PCl5(s) using Hess's Law and given enthalpy data. Includes step-by-step manipulation of reactions and calculations, ideal for college-level chemistry.

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