Q4. The energy of combustion of octane measured in a bomb calorimeter is ΔcU = −5459 kJ mol−1 at 298 K. Find ΔcH for C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l).
Δng = 8 − 12.5 = −4.5. ΔH = ΔU + ΔngRT = −5459 + (−4.5 × 8.314 × 10−3 × 298) = −5459 − 11.1, about −5470 kJ mol−1.
Q5. Why is a calorimeter stirred and insulated?
Stirring spreads the heat evenly, so the thermometer reads the temperature of the whole liquid. Insulation reduces heat leakage to or from the room, so the temperature change closely reflects the reaction heat; accurate work corrects residual heat leakage.
Q6. During an exothermic reaction in a coffee-cup calorimeter some heat leaks out through the lid. Will the measured ΔrH be too negative or not negative enough?
Not negative enough. The lost heat never warms the solution, so ΔT, and hence the size of q, is too small; ΔrH comes out less negative than the true value.
Q7. In the calorimeter of NCERT Problem 5.6 (heat capacity 20.7 kJ K−1), a reaction raises the temperature by 0.50 K. What heat did the reaction release?
q = −CΔT = −20.7 × 0.50, −10.35 kJ before rounding, or about −1.0 × 101 kJ to two significant figures: about 10 kJ was released.
Q8. 0.160 g of methane (M = 16.0 g mol−1) is burnt in the bomb calorimeter of NCERT Problem 5.6 (heat capacity 20.7 kJ K−1), and the temperature rises by 0.428 K. Find ΔcU and ΔcH at 298 K for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l).
Heat gained by the calorimeter = 20.7 × 0.428 = 8.86 kJ for 0.0100 mol, so ΔcU = −886 kJ mol−1. Δng = 1 − 3 = −2, so ΔcH = −886 − 2 × 8.314 × 10−3 × 298 = −886 − 4.96, about −891 kJ mol−1 (NCERT Exercise 5.5 gives −890.3).