The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from 1kg of coke is (assume coke to be 100% carbon). (Given: Enthalpies of combustion of CO_(2),CO and H_(2) are 393.5 kJ, 285 kJ, 285 kJ respectively all at 298K)

The ratio of heats liberated at 298 K from the combustion of one kg of

1.	The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from 1kg of coke is (assume coke to be 100% carbon). (Given: Enthalpies of combustion of CO_(2),CO and H_(2) are 393.5 kJ, 285 kJ, 285 kJ respectively all at 298K)
Answer» `0.79:1` `0.69:1` `0.86:1` `0.96:1` Solution :Number of moles in 1 kg coke`=(1000)/(12)=83.33mol` For the combustion of 1 kg of coke `C+O_(2) to CO_(2)""DeltaH=393.5kJ` `implies`HEAT liberated from 1 mole of coke=393.5 kJ `therefore`Heat liberated from 83.33 mole of coke `H_(1)=(393.5xx83.33)kJ` Also, for the BURNING of water gas `C+H_(2)O to underset("Water gas")ubrace(CO+H_(2))` `CO+H_(2)+O_(2)toCO_(2)+H_(2)O,DeltaH=285+285=570kJ` `therefore`Ratio of heat liberated from burning of water gas obtained from 1 mole of coke `H_(2)=(570xx83.33)kJ` `therefore`REQUIRED ratio `=H_(1):H_(2)=393.5:570=0.69:1`