## Calculate the heat liberated by the combustion of one mole of propane at 298K and 1 atm.

1.The enthalpy change associated with freeze of water at 273K is -6.OkJmol”. The heat capacity (CJ for water is 75.3JmoP’R’ and for ice 37,6Jmol-‘K-‘. Calculate the enthalpy change when water fieezes at 263K.

2. The enthalpy changes at 298K and latm for the hydrogenation and for the combustion of propane are given below : C,&O + HZO = C3Hsk) C3Hs(g) + 502(g) = 3COZ(g) + 4HzO(I) AHI = -124kJmol-‘ AHz=-2,220kJmol~’ In addition the enthalpy change at 298K and latm of the following reaction is known : H20 + 1/2020 = HZO(C) A& = -286kJmoI’ Calculate the heat liberated by the combustion of one mole of propane at 298K and 1 atm.

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