The ratio of heats liberated at ( 298 ~K ) from the combustion of one ( kg ) of coke and by burning water gas obtained from ( 1 ~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 all at ( 298 ) K.)

Step-by-step solution

One kg of coke = 1000 12 =83.33 moles of carbon By burning of one kg of coke C + O ₂ CO ₂ ; _ c H =83.33 393.5 ~kJ (1) Coke By burning of water gas so obtain C + H ₂ O CO + H ₂ CO + H ₂+ O ₂ CO ₂+ H ₂ O ; _ c H =83.33 283.5 ~kJ +83.33 285.5 ~kJ (2)=83.33 569 ~kJ Divide Eq. (1) by Eq. (2), we get = 83.33 393.5 ~kJ 83.33 569 ~kJ =0.69: 1 Therefore, the ratio of heat liberated is 0.69: 1