Enthalpy Formation from Enthalpy Reaction

Question

A scientist measures the standard enthalpy change for the following reaction to be -14.4 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 HCl(aq) is

Answer 1

ΔH_rxn = ∑nΔH_products - ∑nΔH_reactants

_{This means the overall enthalpy change is equal to the sum of the enthalpies of formation of products (times the number of moles) MINUS the sum of the enthalpies of formation of the reactants (times the number of moles).}

Using the values of enthalpies of formation from an appendix in the book:

-14.4 kJ = [(1mol_CaCl2)(-795.5 kJ/mol)+(2 mol_H2O)(-285.8 kJ/mol)] -[(1 mol_Ca(OH)2)(-1002.82 kJ/mol) + (2 mol_HCl)(x)]

-14.4 kJ = [-795.5 kJ - 571.6 kJ] - [-1002.8 kJ + 2mol (x)]

-14.4 kJ = -1367.1 kJ + 1002.8 kJ - 2mol (x)

349.9 kJ = -2mol (x)

-174.95 kJ/mol = x

*NOTE: There may be some discrepancy with these values, as they can vary between textbooks. For accurate calculations, please use the resources your instructor has provided for enthalpies of formation values.*