To determine the enthalpy of the given overall chemical reaction [tex]\( PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g) \)[/tex], we need to use the provided intermediate reactions and their enthalpies. Let's proceed step by step.
### Step 1: Analyze the Given Intermediate Reactions
The given intermediate chemical reactions are:
1. [tex]\( P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(g) \)[/tex]
[tex]\[ \Delta H_1 = -2439 \, \text{kJ} \][/tex]
2. [tex]\( 4PCl_5(g) \rightarrow P_4(s) + 10Cl_2(g) \)[/tex]
[tex]\[ \Delta H_2 = 3438 \, \text{kJ} \][/tex]
### Step 2: Reverse the Reactions as Needed
To form the target reaction, we might need to reverse the given reactions. When a reaction is reversed, the sign of the enthalpy change should also be reversed.
#### Reverse Reaction 1:
[tex]\( 4PCl_3(g) \rightarrow P_4(s) + 6Cl_2(g) \)[/tex]
[tex]\[ \Delta H = +2439 \, \text{kJ} \][/tex]
#### Reaction 2 (as-is):
[tex]\( P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(g) \)[/tex]
[tex]\[ \Delta H = -3438 \, \text{kJ} \][/tex]
### Step 3: Combine the Reactions to Derive the Target Reaction
We add the reversed Reaction 1 and Reaction 2:
[tex]\[ \begin{array}{l}
4PCl_3(g) \rightarrow P_4(s) + 6Cl_2(g) \quad (\Delta H = +2439 \, \text{kJ}) \\
P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(g) \quad (\Delta H = -3438 \, \text{kJ}) \\
\end{array} \][/tex]
Combine the two reactions:
[tex]\[ 4PCl_3(g) + 4Cl_2(g) \rightarrow 4PCl_5(g) \][/tex]
### Step 4: Simplify the Combined Reaction
To match the coefficients of the desired reaction, we can divide the entire equation by 4:
[tex]\[ PCl_3(g) + Cl_2(g) \rightarrow PCl_5(g) \][/tex]
### Step 5: Adjust to the Desired Reaction Form and Calculate Enthalpy
Our target reaction is [tex]\( PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g) \)[/tex]. We reverse the simplified combined reaction and adjust its enthalpy accordingly:
[tex]\[ PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g) \][/tex]
The enthalpy change for this reversed reaction will have the same magnitude but an opposite sign to the enthalpy of the combination we derived:
[tex]\[ \Delta H = \left( \frac{2439 + (-3438)}{4} \right) \text{kJ} \][/tex]
[tex]\[ \Delta H = \frac{-999}{4} \text{kJ} \][/tex]
[tex]\[ \Delta H = -249.75 \text{kJ} \][/tex]
Since the target reaction in the form [tex]\( PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g) \)[/tex] has an enthalpy change of [tex]\( +249.75 \, \text{kJ} \)[/tex] (the magnitude is correct but with positive sign as given in the Python code),
However, the closest options to this calculation given are:
- [tex]\(-250 \, \text{kJ}\)[/tex]
- [tex]\(250 \, \text{kJ}\)[/tex]
### Final Answer
The enthalpy change for the reaction [tex]\( PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g) \)[/tex] is:
[tex]\[ 250 \, \text{kJ} \][/tex]
