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Sagot :
Sure, let's go through the calculation step-by-step for the given reaction, which is:
[tex]\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \][/tex]
We are given the standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) for various substances. The relevant enthalpies for the substances in this reaction are:
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CaCO}_3(s) \)[/tex] = -1206.92 kJ/mol
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CaO}(s) \)[/tex] = -635.09 kJ/mol
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CO}_2(g) \)[/tex] = -393.509 kJ/mol
### Step 1: Calculate the sum of the standard enthalpies of formation for the reactants
For the reactants:
[tex]\[ \sum (\Delta H_{f, \text{reactants}}) = \Delta H_f (\text{CaCO}_3(s)) = -1206.92 \, \text{kJ/mol} \][/tex]
### Step 2: Calculate the sum of the standard enthalpies of formation for the products
For the products:
[tex]\[ \sum (\Delta H_{f, \text{products}}) = \Delta H_f (\text{CaO}(s)) + \Delta H_f (\text{CO}_2(g)) \][/tex]
[tex]\[ \sum (\Delta H_{f, \text{products}}) = -635.09 \, \text{kJ/mol} + (-393.509 \, \text{kJ/mol}) \][/tex]
[tex]\[ = -1028.599 \, \text{kJ/mol} \][/tex]
### Step 3: Use the enthalpy change formula to find the enthalpy of reaction
Using the formula:
[tex]\[ \Delta H_{x \times n} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) \][/tex]
we get:
[tex]\[ \Delta H_{\text{reaction}} = -1028.599 \, \text{kJ/mol} - (-1206.92 \, \text{kJ/mol}) \][/tex]
[tex]\[ = -1028.599 \, \text{kJ/mol} + 1206.92 \, \text{kJ/mol} \][/tex]
[tex]\[ = 178.321 \, \text{kJ/mol} \][/tex]
So, the enthalpy of the reaction is [tex]\( 178.321 \, \text{kJ/mol} \)[/tex].
### Summary:
- Sum of standard enthalpies of formation for reactants: [tex]\( -1206.92 \, \text{kJ/mol} \)[/tex]
- Sum of standard enthalpies of formation for products: [tex]\( -1028.599 \, \text{kJ/mol} \)[/tex]
- Enthalpy of the reaction: [tex]\( 178.321 \, \text{kJ/mol} \)[/tex]
Thus, the enthalpy ([tex]\( \Delta H \)[/tex]) of the reaction [tex]\(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \)[/tex] is [tex]\( 178.321 \, \text{kJ/mol} \)[/tex].
[tex]\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \][/tex]
We are given the standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) for various substances. The relevant enthalpies for the substances in this reaction are:
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CaCO}_3(s) \)[/tex] = -1206.92 kJ/mol
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CaO}(s) \)[/tex] = -635.09 kJ/mol
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CO}_2(g) \)[/tex] = -393.509 kJ/mol
### Step 1: Calculate the sum of the standard enthalpies of formation for the reactants
For the reactants:
[tex]\[ \sum (\Delta H_{f, \text{reactants}}) = \Delta H_f (\text{CaCO}_3(s)) = -1206.92 \, \text{kJ/mol} \][/tex]
### Step 2: Calculate the sum of the standard enthalpies of formation for the products
For the products:
[tex]\[ \sum (\Delta H_{f, \text{products}}) = \Delta H_f (\text{CaO}(s)) + \Delta H_f (\text{CO}_2(g)) \][/tex]
[tex]\[ \sum (\Delta H_{f, \text{products}}) = -635.09 \, \text{kJ/mol} + (-393.509 \, \text{kJ/mol}) \][/tex]
[tex]\[ = -1028.599 \, \text{kJ/mol} \][/tex]
### Step 3: Use the enthalpy change formula to find the enthalpy of reaction
Using the formula:
[tex]\[ \Delta H_{x \times n} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}}) \][/tex]
we get:
[tex]\[ \Delta H_{\text{reaction}} = -1028.599 \, \text{kJ/mol} - (-1206.92 \, \text{kJ/mol}) \][/tex]
[tex]\[ = -1028.599 \, \text{kJ/mol} + 1206.92 \, \text{kJ/mol} \][/tex]
[tex]\[ = 178.321 \, \text{kJ/mol} \][/tex]
So, the enthalpy of the reaction is [tex]\( 178.321 \, \text{kJ/mol} \)[/tex].
### Summary:
- Sum of standard enthalpies of formation for reactants: [tex]\( -1206.92 \, \text{kJ/mol} \)[/tex]
- Sum of standard enthalpies of formation for products: [tex]\( -1028.599 \, \text{kJ/mol} \)[/tex]
- Enthalpy of the reaction: [tex]\( 178.321 \, \text{kJ/mol} \)[/tex]
Thus, the enthalpy ([tex]\( \Delta H \)[/tex]) of the reaction [tex]\(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \)[/tex] is [tex]\( 178.321 \, \text{kJ/mol} \)[/tex].
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