# Problem: Consider the reactionNa2CO3(s)→Na2O(s)+CO2(g)with the enthalpy of reactionΔH°rxn = 321.5 kJ/molWhat is the enthalpy of formation of Na2O(s)?Express your answer in kilojoules per mole to one decimal place.

###### FREE Expert Solution

Recall that ΔH˚rxn can be calculated from the enthalpy of formation (ΔH˚f) of the reactants and products involved:

$\overline{){\mathbf{∆}}{\mathbf{H}}{{\mathbf{°}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{∆}}{\mathbf{H}}{{\mathbf{°}}}_{\mathbf{f}\mathbf{,}\mathbf{products}}{\mathbf{-}}{\mathbf{∆}}{\mathbf{H}}{{\mathbf{°}}}_{\mathbf{f}\mathbf{,}\mathbf{reactants}}}$

Take note that we need to multiply each ΔH˚by the stoichiometric coefficient. Solving for ΔH˚f:

$\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}{\mathbf{Na}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{-}\mathbf{\left(}\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}}\mathbf{+}\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\right)}$

###### Problem Details

Consider the reaction

Na2CO3(s)→Na2O(s)+CO2(g)

with the enthalpy of reaction

ΔH°rxn = 321.5 kJ/mol

What is the enthalpy of formation of Na2O(s)?