Please help!
Calculate the standard change in enthalpy, ΔH°rxn, for the decomposition of calcium carbonate to calcium oxide
CaCO3 (s) → CaO (s) + CO2 (g) Given that:
ΔH°f CaCO3 (s) = -1206.9 kJ/mol
ΔH°f CaO (s) = -635.1 kJ/mol
ΔH°f CO2 (g) = -393.5 kJ/mol
Hess' Law:
Hr=--635.1 kJ-393.5kJ-(-1205.9)kJ
In my head, looks like about it is endothermic (absorbs heat). I bet that is why limestone has to be heated in a furnace to make quick lime.
To calculate the standard change in enthalpy (ΔH°rxn) for the decomposition of calcium carbonate to calcium oxide, you can use the following equation:
ΔH°rxn = ΣΔH°f (products) - ΣΔH°f (reactants)
Let's calculate step-by-step:
1. Write out the balanced equation for the reaction:
CaCO3 (s) → CaO (s) + CO2 (g)
2. Determine the stoichiometric coefficients for each compound:
CaCO3 (s): 1 mole
CaO (s): 1 mole
CO2 (g): 1 mole
3. Plug in the values for the standard heat of formations (ΔH°f) for each compound:
ΔH°f CaCO3 (s) = -1206.9 kJ/mol
ΔH°f CaO (s) = -635.1 kJ/mol
ΔH°f CO2 (g) = -393.5 kJ/mol
4. Calculate the standard change in enthalpy (ΔH°rxn):
ΔH°rxn = [1 × (ΔH°f CaO) + 1 × (ΔH°f CO2)] - [1 × (ΔH°f CaCO3)]
ΔH°rxn = [1 × (-635.1) + 1 × (-393.5)] - [1 × (-1206.9)]
ΔH°rxn = -1028.6 kJ/mol
Therefore, the standard change in enthalpy (ΔH°rxn) for the decomposition of calcium carbonate to calcium oxide is -1028.6 kJ/mol.
To calculate the standard change in enthalpy, ΔH°rxn, for the decomposition of calcium carbonate to calcium oxide, you need to use the Hess's law of heat summation. Hess's law states that the total enthalpy change of a reaction is independent of the intermediate steps and depends only on the initial and final conditions.
First, you need to write the balanced chemical equation for the reaction:
CaCO3 (s) → CaO (s) + CO2 (g)
Now, calculate the ΔH°rxn using the following formula:
ΔH°rxn = ΣΔH°f (products) - ΣΔH°f (reactants)
Where:
ΣΔH°f (products) = sum of the standard enthalpies of formation of the products
ΣΔH°f (reactants) = sum of the standard enthalpies of formation of the reactants
Now, substitute the given values:
ΔH°rxn = [ΔH°f CaO (s) + ΔH°f CO2 (g)] - ΔH°f CaCO3 (s)
ΔH°rxn = [-635.1 kJ/mol + (-393.5 kJ/mol)] - (-1206.9 kJ/mol)
ΔH°rxn = -1028.6 kJ/mol +1206.9 kJ/mol
ΔH°rxn = 178.3 kJ/mol
Therefore, the standard change in enthalpy, ΔH°rxn, for the decomposition of calcium carbonate to calcium oxide is 178.3 kJ/mol.