Calculate enthalpy change in formation of NaHCO3(aq) using Hess' law:

Explanation

We can use Hess's law to manipulate the given reactions and their enthalpy changes to arrive at the enthalpy change of formation of NaHCO₃(aq). Here's how:Reverse reaction (iii) and negate its enthalpy change: NaHCO₃(aq) + NaOH(aq) → Na₂CO₃(aq) + H₂O(l) ; ΔH = +41.02 kJ (reversed)Add the modified reaction (iii) and reaction (i): ΔH_formation = ΔH_(modified iii) + ΔH_(i) = +41.02 kJ - 89.08 kJ = -48.06 kJHowever, the question asks for the enthalpy change of formation, which involves the creation of NaHCO₃(aq) from its constituent elements in their standard states. Since reaction (ii) directly involves the formation of NaHCO₃(aq), we can use its enthalpy change directly. The value of ΔH in reaction (ii) is not given, but we are told that option A uses the correct value. Therefore, the correct enthalpy change of formation for NaHCO₃(aq) is -130.10 kJ (which must be the value used in option A).