To calculate the standard enthalpy change (ΔH⁰) for the reaction between solid sodium hydroxide (NaOH) and gaseous hydrochloric acid (HCl), we will need the standard enthalpies of formation for each reactant. The enthalpy change for the reaction can be calculated using the formula:
ΔH⁰ = ΣΔH⁰ (products) – ΣΔH⁰ (reactants)
First, we will find the standard enthalpies of formation from the provided table:
- ΔH⁰ (NaOH, s) = -425.6 kJ/mol
- ΔH⁰ (HCl, g) = -92.3 kJ/mol
Next, we look at the products of the reaction. In this case, the products are sodium chloride (NaCl) and water (H₂O). Therefore, we need the standard enthalpies for these compounds as well:
- ΔH⁰ (NaCl, s) = -411.15 kJ/mol
- ΔH⁰ (H₂O, l) = -285.83 kJ/mol
Now, we can substitute these values into the formula:
ΔH⁰ = [ΔH⁰ (NaCl) + ΔH⁰ (H₂O)] – [ΔH⁰ (NaOH) + ΔH⁰ (HCl)]
ΔH⁰ = [(-411.15) + (-285.83)] – [(-425.6) + (-92.3)]
ΔH⁰ = -696.98 – (-517.9)
ΔH⁰ = -696.98 + 517.9
ΔH⁰ = -179.08 kJ/mol
Thus, the standard enthalpy change for the reaction is approximately -179.08 kJ/mol. This negative value indicates that the reaction is exothermic, meaning it releases heat.