The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning watergas obtained from kg of coke is (Assume coke to be 100% carbon). (Given enthalpies of combustion of CO_2, CO and H_2 as 393.5 kJ, 285 kJ, 285 kJ respectively at 298 K).

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 watergas obtained from kg of coke is (Assume coke to be 100% carbon). (Given enthalpies of combustion of CO_2, CO and H_2 as 393.5 kJ, 285 kJ, 285 kJ respectively at 298 K).
Answer» `0.79 : 1` `0.69 : 1` `0.86 : 1` `0.96 : 1` Solution :1 kg of coke `= (1000g)/(12 g mol^(-1)) = 83.33 mol` `C + O_(2) rarr CO_(2)` Heat liberated by BURNING 83.33 mol of coke `= 83.33 xx 393.5 kJ` `ubrace(CO + H_(2))_("water GAS") + O_(2) rarr CO_(2) + H_(2)O` Heat liberated by burning 1 mol of water gas `= 285 + 285 = 570 kJ`. Heat liberated by burning water gas obtained from 1 kg of coke `= 83.33 xx 570 kJ` Ratio of heat liberated Coke : water gas `83.33 xx 393.5 : 83.33 xx 570` `393.5 : 570` `0.69 : 1`