Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
