Use this information and the additional information provided to calculate the ΔH°f of NaHCO3.

Question

When 39.0 g NaHCO₃ decomposes, the change in energy is-30.0 kJ. Use this information and the additional information provided to calculate the ΔH°_f of NaHCO₃.

The H_f values are given below

H₂O(l) = -286

CO₂(g) = -394

Na₂CO₃(s) = -1131

Equation: NaHCO₃ (s) ---> H₂O(l) + Na₂CO₃(s) + CO₂(g)

I've calculated the H_f of the products side ( -2942). I'm confused on what to do after this point and if I should divide my final value by two for one mole.

Answer 1

2NaHCO₃ (s) ---> H₂O(l) + Na₂CO₃(s) + CO₂(g) ... balanced equation

Not sure how you got -2942 for H_f of products, but anyway....

The ∆H_rxn = ∆H_f products - ∆H_f reactants

We can find the ∆H_rxn from the given info that 39.0 g NaHCO₃ produces 30.0 kJ of heat. We will convert this to heat generated per 1 mole of NaHCO₃:

39.0 g NaHCO₃ x 1 mol NaHCO₃ / 84.0 g = 0.46 mols NaHCO₃

-30.0 kJ / 0.46 mol = -65.2 kJ/mol = ∆H_rxn (per mole)

∆H_rxn = ∆H_f products - ∆H_f reactants

-65.2 = (-286 + -1131 + -394) - 2x∆H_f NaHCO₃

-65.2 = -1811 - 2x∆H_f NaHCO₃

∆H_f NaHCO₃ = -1746 kJ/2

∆H_f NaHCO3 = -873 kJ/mole