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:

\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.