Question

Calculate the change in enthalpy (in kJ/mol) for the following reaction using the data provided.

2NaHCO3(s) --> Na2CO3(s) + H2O(l) + CO2(g) ΔH°f = ?

ΔH°f = NaHCO3(s) = -947.7 kJ/mol

ΔH°f = Na2CO3(s) = -1131 kJ/mol

ΔH°f = H2O(g) = -241.8 kJ/mol

ΔH°f = H2O(l) = -285.9 kJ/mol

ΔH°f = CO2(g) = -393.5 kJ/mol

Answer 1

Solution:

Since the enthalpy changes for reactions are the standard enthalpies of formation, it follows

∆H = ∑ ∆Hf^o(products) – ∑ ∆Hf^o(reactants)

For the following reaction the change in enthalpy can be calculate:

ΔH = [∆Hf^o Na₂CO₃ + ΔH°f H₂O(l) + ΔH°f CO₂(g)] – [2×ΔH°f NaHCO₃] = [(-1131) + (-285.9) + (-393.5)] – [2×(-947.7)] = 85 kJ/mol

Answer: the change in enthalpy ΔH 85 kJ/mol.