From simple questions to complex issues, IDNLearn.com has the answers you need. Our community is here to provide detailed and trustworthy answers to any questions you may have.
Sagot :
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].
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.
