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To determine the molar enthalpy change for the reaction [tex]\( C(s) + 4 \text{HF}(g) \rightarrow \text{CF}_4(g) + 2 \text{H}_2(g) \)[/tex], we will make use of two given reactions and their respective enthalpy changes ([tex]\(\Delta H\)[/tex]):
1. [tex]\( \text{H}_2(g) + \text{F}_2(g) \rightarrow 2 \text{HF}(g) \)[/tex], [tex]\( \Delta H^\circ = -79.2 \, \text{kJ/mol} \)[/tex]
2. [tex]\( \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \)[/tex], [tex]\( \Delta H^\circ = 141.3 \, \text{kJ/mol} \)[/tex]
Let’s analyze the given reactions step-by-step:
### Step 1: Reverse the first reaction
The first reaction (formation of 2 HF from H2 and F2) needs to be reversed, as we have 4 HF on the reactant side in the target reaction. Reversing the direction of a chemical equation inverts the sign of [tex]\(\Delta H\)[/tex]:
[tex]\[ 2 \text{HF}(g) \rightarrow \text{H}_2(g) + \text{F}_2(g) \][/tex]
[tex]\[ \Delta H^\circ = +79.2 \, \text{kJ/mol} \][/tex]
### Step 2: Double the reversed reaction 1
To match the 4 HF from the target reaction, we need to multiply the reversed reaction by 2:
[tex]\[ 4 \text{HF}(g) \rightarrow 2 \text{H}_2(g) + 2 \text{F}_2(g) \][/tex]
[tex]\[ \Delta H^\circ = 2 \times 79.2\, \text{kJ/mol} = 158.4 \, \text{kJ/mol} \][/tex]
### Step 3: Combine with reaction 2
Now, consider reaction 2:
[tex]\[ \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \][/tex]
[tex]\[ \Delta H^\circ = 141.3 \, \text{kJ/mol} \][/tex]
### Step 4: Combine the adjusted reactions
We need to add the doubled reversed reaction 1 to reaction 2. The combination looks like this:
[tex]\[ \left( 4 \text{HF}(g) \rightarrow 2 \text{H}_2(g) + 2 \text{F}_2(g) \right) + \left( \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \right) \][/tex]
By combining:
[tex]\[ 4 \text{HF}(g) + \text{C}(s) + 2 \text{F}_2(g) \rightarrow 2 \text{H}_2(g) + 2 \text{F}_2(g) + \text{CF}_4(g) \][/tex]
### Step 5: Simplify the overall equation
The 2 \text{F}_2(g) on both sides cancel out:
[tex]\[ \text{C}(s) + 4 \text{HF}(g) \rightarrow \text{CF}_4(g) + 2 \text{H}_2(g) \][/tex]
This is exactly the target reaction.
### Step 6: Calculate the total molar enthalpy change
The total enthalpy change for the combined reactions is the sum of their individual enthalpy changes:
[tex]\[ \Delta H_{\text{total}} = 158.4 \, \text{kJ/mol} + 141.3 \, \text{kJ/mol} = 299.7 \, \text{kJ/mol} \][/tex]
Therefore, the molar enthalpy change for the reaction [tex]\( \text{C}(s) + 4 \text{HF}(g) \rightarrow \text{CF}_4(g) + 2 \text{H}_2(g) \)[/tex] is [tex]\( 299.7 \, \text{kJ/mol} \)[/tex].
1. [tex]\( \text{H}_2(g) + \text{F}_2(g) \rightarrow 2 \text{HF}(g) \)[/tex], [tex]\( \Delta H^\circ = -79.2 \, \text{kJ/mol} \)[/tex]
2. [tex]\( \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \)[/tex], [tex]\( \Delta H^\circ = 141.3 \, \text{kJ/mol} \)[/tex]
Let’s analyze the given reactions step-by-step:
### Step 1: Reverse the first reaction
The first reaction (formation of 2 HF from H2 and F2) needs to be reversed, as we have 4 HF on the reactant side in the target reaction. Reversing the direction of a chemical equation inverts the sign of [tex]\(\Delta H\)[/tex]:
[tex]\[ 2 \text{HF}(g) \rightarrow \text{H}_2(g) + \text{F}_2(g) \][/tex]
[tex]\[ \Delta H^\circ = +79.2 \, \text{kJ/mol} \][/tex]
### Step 2: Double the reversed reaction 1
To match the 4 HF from the target reaction, we need to multiply the reversed reaction by 2:
[tex]\[ 4 \text{HF}(g) \rightarrow 2 \text{H}_2(g) + 2 \text{F}_2(g) \][/tex]
[tex]\[ \Delta H^\circ = 2 \times 79.2\, \text{kJ/mol} = 158.4 \, \text{kJ/mol} \][/tex]
### Step 3: Combine with reaction 2
Now, consider reaction 2:
[tex]\[ \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \][/tex]
[tex]\[ \Delta H^\circ = 141.3 \, \text{kJ/mol} \][/tex]
### Step 4: Combine the adjusted reactions
We need to add the doubled reversed reaction 1 to reaction 2. The combination looks like this:
[tex]\[ \left( 4 \text{HF}(g) \rightarrow 2 \text{H}_2(g) + 2 \text{F}_2(g) \right) + \left( \text{C}(s) + 2 \text{F}_2(g) \rightarrow \text{CF}_4(g) \right) \][/tex]
By combining:
[tex]\[ 4 \text{HF}(g) + \text{C}(s) + 2 \text{F}_2(g) \rightarrow 2 \text{H}_2(g) + 2 \text{F}_2(g) + \text{CF}_4(g) \][/tex]
### Step 5: Simplify the overall equation
The 2 \text{F}_2(g) on both sides cancel out:
[tex]\[ \text{C}(s) + 4 \text{HF}(g) \rightarrow \text{CF}_4(g) + 2 \text{H}_2(g) \][/tex]
This is exactly the target reaction.
### Step 6: Calculate the total molar enthalpy change
The total enthalpy change for the combined reactions is the sum of their individual enthalpy changes:
[tex]\[ \Delta H_{\text{total}} = 158.4 \, \text{kJ/mol} + 141.3 \, \text{kJ/mol} = 299.7 \, \text{kJ/mol} \][/tex]
Therefore, the molar enthalpy change for the reaction [tex]\( \text{C}(s) + 4 \text{HF}(g) \rightarrow \text{CF}_4(g) + 2 \text{H}_2(g) \)[/tex] is [tex]\( 299.7 \, \text{kJ/mol} \)[/tex].
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