Astudent mixed 25.0cm3 of 0.350 moldm–3 sodium hydroxide solution with 25.0cm3 of 0.350moldm–3 hydrochloric acid. the temperature rose by 2.50°c. assume that no heat was lost to the surroundings. the final mixture had a specific heat capacity of 4.20j cm–3k–1. what is the molar enthalpy change for the reaction? a –150kj mol–1 b –60.0kj mol–1 c –30.0kj mol–1 d –0.150kj mol–1

Step 1: Enthalpy change for the reaction.

Total volume of water = 25 cm3 + 25 cm3 = 50 cm3 = 0.05 kg

Now, change in enthalpy = ΔH = cm ΔT

                                        = 4.20 × 0.05 × 2.5

                                        = -0.525 kJ

Since, the temperature has increased, enthalpy change is negative.

Step 2: Calculating the number of moles of water produced.

From the balanced equation, it can be seen that the number of moles of water is equal to the number of moles of nitric acid:

HCl(aq) + NaOH(aq)    →            NaCl(aq) + H2O(l)

Number of moles of HCl = concentration × volume (litres)

                                       = 0.350 × 0.025

                                       = 0.00875 mole

Step 3: Calculating the enthalpy change for one mole of water.

ΔH = -0.525 kJ, for 0.00875 mole of water.

Thus,    0.00875   mole   ≡   -0.525 kJ

                ∴1         mole   ≡       -60 kJ

Thus, the molar enthalpy change for the reaction is -60 kJ/mol