Using standard heats of formation, calculate the standard enthalpy change for the following reaction. Ca(OH)2(aq) 2HCl(aq)CaCl2(s) 2H2O(l)

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)

false

because it does not have the potential to be entirely used up

answer: i dont know the answer but this might

explanation:

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