Question

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.

Answer 1

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.