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To determine the enthalpy change ([tex]\( \Delta H \)[/tex]) for the overall reaction, we need to manipulate the given intermediate reactions to match the final reaction. The overall reaction we are aiming for is:
[tex]\[ \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) \][/tex]
Let's examine the given intermediate reactions:
1. [tex]\[ \text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s) \quad \Delta H_1 = -812.8 \, \text{kJ} \][/tex]
2. [tex]\[ 2 \text{Ca} (s) + \text{O}_2 (g) \rightarrow 2 \text{CaO} (s) \quad \Delta H_2 = -1289.8 \, \text{kJ} \][/tex]
For the first intermediate reaction, we notice that it already contains [tex]\(\text{CaCO}_3 (s)\)[/tex] on the product side, which is part of our desired overall reaction. So we can use it as is.
For the second intermediate reaction, we have 2 moles of Ca and 2 moles of CaO. We want to only have 1 mole of CaO, so we will need to reverse half of this reaction to match our desired overall reaction. Here's how we do it:
First, we reverse the reaction:
[tex]\[ 2 \text{CaO} (s) \rightarrow 2 \text{Ca} (s) + \text{O}_2 (g) \][/tex]
When we reverse a reaction, the enthalpy change sign is reversed:
[tex]\[ \Delta H = +1289.8 \, \text{kJ}\][/tex]
Next, we halve the reaction to get:
[tex]\[ \text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g) \][/tex]
Halving the reaction means halving the enthalpy change:
[tex]\[ \Delta H = \frac{1289.8}{2} = 644.9 \, \text{kJ} \][/tex]
Now, we combine the two reactions:
Intermediate reactions:
[tex]\[ \text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s) \quad \Delta H_1 = -812.8 \, \text{kJ} \][/tex]
[tex]\[ \text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g) \quad \Delta H = 644.9 \, \text{kJ} \][/tex]
When we add these intermediate reactions together:
[tex]\[ (\text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s)) + (\text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g)) \][/tex]
The result is:
[tex]\[ \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s)\][/tex]
The enthalpy change for the overall reaction is:
[tex]\[ \Delta H_{\text{overall}} = -812.8 \, \text{kJ} + 644.9 \, \text{kJ} = -167.9 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the overall reaction [tex]\( \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) \)[/tex] is [tex]\(-167.9 \, \text{kJ}\)[/tex].
[tex]\[ \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) \][/tex]
Let's examine the given intermediate reactions:
1. [tex]\[ \text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s) \quad \Delta H_1 = -812.8 \, \text{kJ} \][/tex]
2. [tex]\[ 2 \text{Ca} (s) + \text{O}_2 (g) \rightarrow 2 \text{CaO} (s) \quad \Delta H_2 = -1289.8 \, \text{kJ} \][/tex]
For the first intermediate reaction, we notice that it already contains [tex]\(\text{CaCO}_3 (s)\)[/tex] on the product side, which is part of our desired overall reaction. So we can use it as is.
For the second intermediate reaction, we have 2 moles of Ca and 2 moles of CaO. We want to only have 1 mole of CaO, so we will need to reverse half of this reaction to match our desired overall reaction. Here's how we do it:
First, we reverse the reaction:
[tex]\[ 2 \text{CaO} (s) \rightarrow 2 \text{Ca} (s) + \text{O}_2 (g) \][/tex]
When we reverse a reaction, the enthalpy change sign is reversed:
[tex]\[ \Delta H = +1289.8 \, \text{kJ}\][/tex]
Next, we halve the reaction to get:
[tex]\[ \text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g) \][/tex]
Halving the reaction means halving the enthalpy change:
[tex]\[ \Delta H = \frac{1289.8}{2} = 644.9 \, \text{kJ} \][/tex]
Now, we combine the two reactions:
Intermediate reactions:
[tex]\[ \text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s) \quad \Delta H_1 = -812.8 \, \text{kJ} \][/tex]
[tex]\[ \text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g) \quad \Delta H = 644.9 \, \text{kJ} \][/tex]
When we add these intermediate reactions together:
[tex]\[ (\text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s)) + (\text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g)) \][/tex]
The result is:
[tex]\[ \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s)\][/tex]
The enthalpy change for the overall reaction is:
[tex]\[ \Delta H_{\text{overall}} = -812.8 \, \text{kJ} + 644.9 \, \text{kJ} = -167.9 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the overall reaction [tex]\( \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) \)[/tex] is [tex]\(-167.9 \, \text{kJ}\)[/tex].
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