HCl + NaHCO3 + NaCl + H2O + CO2

The enthalpy change for this reaction can be calculated using the following equation:

ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)

where ΔH is the enthalpy change, n is the number of moles of each substance, and ΔHf is the standard enthalpy of formation of each substance.

The standard enthalpies of formation for the substances involved in the reaction are:

HCl: -167.2 kJ/mol

NaHCO3: -950.8 kJ/mol

NaCl: -411.2 kJ/mol

H2O: -285.8 kJ/mol

CO2: -393.5 kJ/mol

Substituting these values into the equation, we get:

ΔH = [1(-393.5 kJ/mol) + 1(-411.2 kJ/mol)] - [1(-167.2 kJ/mol) + 1(-950.8 kJ/mol)]

ΔH = -804.7 kJ/mol + 1118 kJ/mol

ΔH = 313.3 kJ/mol

Therefore, the enthalpy change for the reaction is 313.3 kJ/mol. This means that the reaction is endothermic, meaning that it absorbs heat from the surroundings.