Answer:
ΔH = -29.78 KJ/mol
Explanation:
To calculate the Standard enthalpy of change for this reaction, we need to use the following expression:
ΔH = ΔHf(products) - ΔHf(reactants) (1)
Now, the standard heat of formation can be found in different handbooks or websites. I manage to find these for the compounds of the reaction:
ΔHf Ca(OH)₂ = -1002.82 KJ/mol
ΔHf HCl = -167.4 KJ/mol
ΔHf CaCl₂ = -795.8 KJ/mol
ΔHf H₂O = -285.8 KJ/mol
With these values, we can calculate the enthalpy change of the reaction using (1). Let's calculate first the individual values of the products and reactants and then, we can use (1):
Reaction: Ca(OH)₂ + 2HCl ---------> CaCl₂ + 2H₂O
ΔHf (products) = ΔHf CaCl₂ + ΔHf H₂O
ΔHf (products) = (-795.8) + 2(-285.8)
ΔHf (products) = -1367.4 KJ/mol
Keep in mind that if the compound has a coefficient that balances the reaction, you need to multiply that number by the standard heat. Let's calculate reactants:
ΔHf (reactants) = ΔHf Ca(OH)₂ + ΔHf HCl
ΔHf (reactants) = (-1002.82) + 2(-167.4)
ΔHf (reactants) = -1337.62 KJ/mol
Finally using expression (1) we have:
ΔH = (-1367.4) - (-1337.22)
