The balanced equation for the decomposition of sodium bicarbonate is 2 NaHCO3 (s) --> Na2CO3 (s) + H2O (g) + CO2 (g)delta H 129.2. Using the enthalpies of formation, calculate the enthalpy of formation of sodium bicarbonate.

dHrxn = (n*dHf products) - (n*dHf reactants)

dHrxn = 129.2 kJ
Substitute the others and solve for dH NaHCO3.

To calculate the enthalpy of formation of sodium bicarbonate (NaHCO3), we need to use the enthalpies of formation of the reactants and products involved in the balanced equation. The enthalpy of formation (∆Hf) is defined as the change in enthalpy that occurs when one mole of a compound is formed from its elements in their standard states at a given temperature and pressure (usually 298 K and 1 atm).

The balanced equation for the decomposition of sodium bicarbonate is:
2 NaHCO3 (s) --> Na2CO3 (s) + H2O (g) + CO2 (g) (∆H = 129.2 kJ)

To calculate the ∆Hf of NaHCO3, we will use the enthalpies of formation for Na2CO3, H2O, and CO2, as well as their stoichiometric coefficients from the balanced equation.

The enthalpies of formation (∆Hf) at 298 K and 1 atm are given as:

∆Hf° of Na2CO3 (s) = -1130.0 kJ/mol
∆Hf° of H2O (g) = -241.8 kJ/mol
∆Hf° of CO2 (g) = -393.5 kJ/mol

Using Hess's Law, we can sum the enthalpies of formation of the products and subtract the enthalpies of formation of the reactants to calculate the enthalpy of formation of NaHCO3.

∆Hf° of NaHCO3 = [(1 mol Na2CO3) * (-1130.0 kJ/mol)] + [(1 mol H2O) * (-241.8 kJ/mol)] + [(1 mol CO2) * (-393.5 kJ/mol')] - [(2 mol NaHCO3) * (∆H = 129.2 kJ)]

∆Hf° of NaHCO3 = (-1130.0 kJ/mol) + (-241.8 kJ/mol) + (-393.5 kJ/mol) - [(2 mol) * (129.2 kJ/mol)]

Make sure to calculate the right side of the equation before subtracting from the left side. By performing the calculation, you should get the value for the enthalpy of formation of sodium bicarbonate (∆Hf° of NaHCO3).