The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas 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).

Options

A. `0.79 : 1`
B. `0.69 : 1`
C. `0.86 : 1`
D. `0.96 : 1`

Answer

B

Explanation

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`