Answer to Question #56366 in Physical Chemistry for Alison
In an experiment to determine the enthalpy change (△H) for the neutralisation reaction
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
100 mL (VHCl) of a 0.5 M (CHCl) HCl solution are reacted with 100 mL (VNaOH) of a 0.5 M (CNaOH) NaOH solution. The initial temperature of the system (Ti) is 20.1°C, whereas the final temperature (Tf) is 23.4°C. Assume that the density (p) of both 0.5 M HCl and 0.5 M NaOH solutions is 1.0 g mL-1, and that the specific heat capacity (Cp) is 4.183 J g-1 K-1.
a) Calculate the amount of heat in J (i.e. the enthalpy change △H) involved in the reaction, stating whether it is endothermic or exothermic.
b) Calculate the molar enthalpy in kJ mol-1 (Hm) for the reaction.
a) We will calculate amount of heat using: Q = Cp · m · ΔT, where Ср – the specific heat, J/(g·K); m – mass of solution; ΔT – change of temperature, K. Having substituted data from a statement of the problem in this formula, we will receive: Q = 4,183·(1,0*100 +1,0*100) · (23,4 – 20,1) = 2760,78 J. ΔН = – Q = –2760,78 J < 0 – reaction is exothermic.
b) Number of moles of sodium hydroxide equals to number of moles of HCl: νNaOH = νHCl = 0,5*100/1000 = 0.05 mol Then the molar enthalpy in kJ mol-1 ( Hm) for the reaction is equal: Hm = – Q = –2760,78/0,05 = –55215,6 J/mol = –55,22 kJ/mol.
Answer: a) ΔН =–2760,78 J < 0 – reaction is exothermic; b) Hm = –55,22 kJ/mol.