Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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]
### 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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
