Respuesta :
Answer:
The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ
Explanation:
Step 1: Data given
temperature = 850 °C
Mass of 68.10 grams of CO2
ΔH°f (CaO) = -635.6 kJ/mol
ΔH°f (CO2) = -693.5 kJ/mol
ΔH°f (CaCO3) =-1206.9 kJ/mol
Step 2: The balanced equation
CaCO3(s) → CaO(s) + CO2(g)
Step 3: Calculate ΔH°reaction
ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)
ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) - ΔH°f (CaCO3)
ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol
ΔH°reaction = -122.2 kJ /mol
Step 4: Calculate moles of CO2
Moles CO2 = mass CO2 / Molar mass CO2
Moles CO2 = 68.10 grams / 44.01 g/mol
Moles CO2 = 1.547 moles
Step 5: Calculate the enthalpy change for 68.10 grams of CO2
-122.2 kJ/mol * 1.547 moles = -189.04 kJ
The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ
The enthalpy change when 68.10 g of CO₂ is produced in the decomposition of CaCO₃ is -189.0 kJ.
Let's consider the decomposition of CaCO₃.
CaCO₃(s) ⇒ CaO(s) + CO₂(g)
Given the standard enthalpies of formation (ΔH°f) of products (p) and reactants (r), we can calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.
[tex]\Delta H\°_{rxn} = \Sigma \Delta H\°_f(p) \times n_p - \Sigma \Delta H\°_f(r) \times n_r\\\\\Delta H\°_{rxn} = \Delta H\°_f(CaO) \times 1 mol + \Delta H\°_f(CO_2) \times 1 mol - \Delta H\°_f(CaCO_3) \times 1 mol\\\\\Delta H\°_{rxn} = (-635.6kJ/mol) \times 1 mol + (-693.5kJ/mol) \times 1 mol - (-1206.9 kJ/mol) \times 1 mol = -122.2 kJ[/tex]
The molar mass of CO₂ is 44.01 g/mol. The moles corresponding to 68.10 g are:
[tex]68.10 g \times \frac{1mol}{44.01 g} = 1.547mol[/tex]
122.2 kJ are evolved when 1 mole of CO₂ is formed. The heat evolved when 1.547 moles of CO₂ are evolved is:
[tex]1.547 mol \times \frac{(-122.2 kJ)}{mol} = -189.0 kJ[/tex]
The enthalpy change when 68.10 g of CO₂ is produced in the decomposition of CaCO₃ is -189.0 kJ.
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