Question

25.00 mL of 0.148 M HCl is added to 35.00 mL of 0.162 M NaOH in a coffee cup calorimeter.

The initial temperature of both solutions is 27.50 oC. The temperature of the final mixture is

30.15 oC. Calculate H for the reaction in kJ/mol of water formed. (DH2O = 1.0 g/mL and cH2O

= 4.184 J/g oC)

Answer 1

NaOH + HCl = H2O + NaCl - $\Delta$ H

n(HCl) = 0.025*0.148 = 0.0037 mol

n(NaOH) = 0.035*0.162 = 0.00567 mol

Sodium hydroxide in excess. The following calculations are performed on hydrochloric acid. Find the amount of heat released:

Q = с(H2O)*m(solution)*(t₂-t₁) = 4.181 * (25+35) * (30.15 - 27.5) = 1194 J = 1.194 kJ

Δ H = Q/n(HCl) = 1.194/0.0037 = 322.8 kJ/mol

Answer: Δ H for reaction is 322.8 kJ/mol.