Correct Answer:
Option B i.e. -60 kJ/mol
Reason:
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
