If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The | Hf | of ethanol at 298.15 K is 102 kJ mol-1. (Nearest integer) Given: Standard enthalpy for combustion of graphite = −393.5 kJ mol-1 Standard enthalpy of formation of water (l) = −285.8 kJ mol-1 Molar mass in g mol-1 of C, H, O are 12, 1 and 16 respectively
Answer
3
Solution
Bomb calorimeter (constant volume) ⇒ heat measured = U
n (ethanol) = 3.36546 = 0.07315 mol
⇒ cU = − 99.4720.07315 = −1359.8 kJ mol-1
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ⇒ ng = 2 − 3 = −1
cH = cU + ngRT
= −1359.8 + (−1)(8.314 × 10-3)(298.15)
= −1359.8 − 2.48 = −1362.3 kJ mol-1
cH = 2 fH(CO2) + 3 fH(H2O) - fH(C2H5OH) ( fH(CO2) = cH of graphite)
⇒ −1362.3 = 2(−393.5) + 3(−285.8) − fH
⇒ fH = −787.0 − 857.4 + 1362.3 = −282.1 kJ mol-1
∴ | fH | = 2.82 × 102 kJ mol-1 ≈ 3 × 102 kJ mol-1
Answer: 3