posted by Austin on .
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
Hf(CaCl2) + 2*Hf(H2O)-Hf(Ca(OH2))-2Hf(HCL)=-53.4kJ