A scientist measures the standard enthalpy change for the following reaction to be -53.4 kJ :
Ca(OH)2(aq) + 2 HCl(aq) CaCl2(s) + 2 H2O(l)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is kJ/mol
Hess Law:
Hf(CaCl2) + 2*Hf(H2O)-Hf(Ca(OH2))-2Hf(HCL)=-53.4kJ
To find the standard enthalpy of formation (ΔHf) of H2O(l), we can use the given information about the standard enthalpy change (ΔH) of the reaction and the standard enthalpies of formation (ΔHf) of the other substances involved.
The standard enthalpy change of a reaction is defined as the difference between the standard enthalpies of formation of the products and the reactants:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
In this reaction equation:
Ca(OH)2(aq) + 2 HCl(aq) → CaCl2(s) + 2 H2O(l)
The standard enthalpies of formation for Ca(OH)2(aq) and CaCl2(s) are typically given, and we need to find ΔHf for H2O(l).
Given information: ΔH = -53.4 kJ
Rearranging the equation, we can solve for the unknown ΔHf of H2O(l):
ΔHf H2O(l) = Σ(ΔHf products) - Σ(ΔHf reactants) + ΔH
ΔHf H2O(l) = (ΔHf CaCl2(s)) + 2(ΔHf H2O(l)) - (ΔHf Ca(OH)2(aq)) - 2(ΔHf HCl(aq)) + ΔH
Now, substitute the given values:
ΔHf H2O(l) = 0 + 2(ΔHf H2O(l)) - (ΔHf Ca(OH)2(aq)) - 2(ΔHf HCl(aq)) + (-53.4 kJ)
Simplifying the equation further:
ΔHf H2O(l) - 2(ΔHf H2O(l)) = (-ΔHf Ca(OH)2(aq)) - 2(ΔHf HCl(aq)) - 53.4 kJ
Combine like terms:
-ΔHf H2O(l) = (-ΔHf Ca(OH)2(aq)) - 2(ΔHf HCl(aq)) - 53.4 kJ
Finally, solve for the unknown ΔHf H2O(l):
ΔHf H2O(l) = [(-ΔHf Ca(OH)2(aq)) - 2(ΔHf HCl(aq)) - 53.4 kJ] * (-1)
By substituting the known values for standard enthalpies of formation of Ca(OH)2(aq) and HCl(aq) and solving the equation, we can obtain the value of ΔHf H2O(l).