Answer to Question #113481 in Organic Chemistry for jay mck

Question

When 5.00 g of sodium hydrogen carbonate, NaHCO3(s), reacts completely with 70.00 mL of acetic acid, CH3COOH(aq), the temperature increases from 15.5 C to 22.3C. Assume that the acid solution has the same density and specific heat capacity as water and that the mass of the final solution is 70.00 g. Calculate the molar enthalpy of the reaction, Hr of NaHCO3.

Answer 1

Find the amount of heat released during the reaction:

Q = cm(t₂-t₁) = 4200*0.07*(22.3 - 15.5) = 1999.2 J

Write the equation for the reaction NaHCO₃ and CH₃COOH:

NaHCO₃(s) + CH₃COOH = CH₃COONa + H₂O + CO₂ - $\Delta$ H_r

n(NaHCO₃) = 5/84 = 0.06 mol. Acetic acid in excess. The enthalpy of the reaction is calculated by the number of moles of NaHCO₃.

Q = n(NaHCO₃)*Δ H_r, herefrom:

Δ H_r = Q/n(NaHCO₃) = 1999.2/0.06 = 33320 J/mol = 33.22 kJ/mol

Molar enthalpy of the reaction is 33.22 kJ/mol.