Q4. 50 mL of water at 24.0 °C is in a beaker calorimeter; 50 mL of water at 60.0 °C is poured in and the mixture reaches 41.2 °C. Find the calorimeter constant.
Heat lost by the hot water: 50 × 4.184 × 18.8, about 3933 J. Heat gained by the cold water: 50 × 4.184 × 17.2, about 3598 J. The beaker took about 335 J for a 17.2 K rise, so W = 4.184 × (940 − 860)/17.2, about 19.5 J K−1.
Q5. 50 mL of 1.0 M H2SO4 is mixed with 100 mL of 1.0 M NaOH, both at 25.0 °C, in a calorimeter with W = 34.9 J K−1. Taking ΔneutH = −57 kJ mol−1, predict the final temperature.
Each H2SO4 gives two H+: 0.050 × 2 = 0.100 mol of H+ meets 0.100 mol of OH−, forming 0.100 mol of water. Heat given out = 0.100 × 57 000 = 5700 J. Heat capacity of mixture and beaker = 150 × 4.184 + 34.9, about 662.5 J K−1. Δt = 5700/662.5, about 8.6 K, so the final temperature is about 33.6 °C.
Q6. In the calorimeter of Example 2 (W = 34.9 J K−1), 100 mL of 1.0 M of a weak acid HA and 100 mL of 1.0 M NaOH give a rise of 5.8 K. Find ΔneutH for HA and estimate the enthalpy of ionisation of HA, taking −57.5 kJ mol−1 for a strong acid in this apparatus.
q = (836.8 + 34.9) × 5.8, about 5056 J for 0.100 mol, so ΔneutH is about −50.6 kJ mol−1. By Hess’s law, HA → H+ + A− followed by H+ + OH− → H2O adds up to the neutralisation, so ΔionH = −50.6 − (−57.5), about +6.9 kJ mol−1.
Q7. Why does the ionisation H2O(l) ⇌ H+(aq) + OH−(aq) go further forward when the temperature rises? (A discussion question in the NCERT Laboratory Manual.)
It is the reverse of neutralisation, so it absorbs heat (about +56 kJ mol−1). For an endothermic change, raising the temperature shifts the equilibrium forward, so the ionic product of water increases with temperature in the near-room-temperature range considered here (Unit 6, Equilibrium).
Q8. Why is the measured enthalpy of neutralisation usually a little smaller in size than the true value, and why is the highest temperature after mixing the one recorded?
Some heat leaks out through the lid and insulation while the mixture is still warming up, so the highest reading falls short of the true rise and the heat worked out from it comes out a little low. For a fast reaction and a well-mixed solution, the peak is a useful elementary estimate; thermometer lag and heat loss can lower it. Recording temperature against time and extrapolating the cooling curve back to mixing improves the estimate.