Answer:
-947.6 kJ/mol.
Step-by-step explanation:
The constituent elements combined to result in NaHCO3(s)
NaHCO3(s) → (1/2).Na2CO3(s) + (1/2).CO2(g) + (1/2).H2O(g) : ΔHr = +64.6 kJ/mol
Enthalpy change of a reaction can be measured and given the symbol, . It is equal to the difference in enthalpy between reactants and products
From this equation we deduct that water is produced as a vapour, hence the reaction is endotermic and ΔHr is positive.
Therefore, To find the enthalpy of formation of NaHCO3 from this equation, we use the general equation which relates the enthalpy change ΔHr of the reaction to the enthalpies ΔHf of formation for the reactants and products.
ΔHr = ∑ΔH(products) - ∑ΔH(reagents)
where the
ΔH is equal to the change in the potential energy of enthalpy of the chemical bond.
In ideal test scenarios, these values should be supplied, as you will not be able to look them up. Here, I've supplied values here.
ΔHf : Na2CO3(s) = -1130.7, CO2(g) = -393.5, H2O(g) = -241.8 kJ/mol
Using these figures (you may wish to check them in Wikipedia or elsewhere), we get
+64.6 = [(1/2)*(-1130.7) + (1/2)*(-393.5) + (1/2)*(-241.8)] - ΔHf(NaHCO3)
With the result that the enthalpy of formation of NaHCO3 is obtained as -947.6 kJ/mol.