Calculate the standard enthalpy change for the following reaction at 25 °C.

Question

Calculate the standard enthalpy change for the following reaction at 25 °C.

HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)

Substance

∆Hf °  kJ/mol

∆Gf°  kJ/mol

S°  J/(mol·K

HCl(g)

–92.3

–95.3

186.9

NaOH(s)

–425.8

–379.7

64.4

NaCl(s)

–411.2

–384.1

72.1

H2O(l)

–285.8

–237.1

70.0

Answer 1

You are looking for the ΔH°_rxn value here.

For simplicity, we will call that value H' since there isn't an easy way to type equations on wyzant yet. This is not an official notation, just a simplificaiton for this answer.

The formula we use to find the standard enthalpy change of reaction (ΔHº_rxn) is:

= ∑(n × ΔH°_{f products}) - ∑(m × ΔHº_{f reactants})

n and m are the stoichiometric coefficients of the reaction, which means we need a balanced reaction equation:

HCl + NaOH <-> NaCl + H₂O

We don't need to add any stoichiometric coefficients since this equation is already balanced. None of the reactants or products are elements in their standard state so we will NOT exclude any reactant or product.

Using the formula above and the table given, your standard enthalpy change of reaction is:

= [(1 mol x -411.2 kJ/mol) + (1 mol x -285.8 kJ/mol)] - [(1 mol x -425.8 kJ/mol) + (1 mol x -92.3 kJ/mol)]

= (-411.2 - 285.8) - (-425.8 - 92.3)

= (-697) - (-518.1)

= -697 + 518.1

= -178.9 kJ