The ratio of heats liberated at \( 298 \mathrm{~K} \) from the combustion of one \( \mathrm{kg} \) of coke and by burning
water gas obtained from \( 1 \mathrm{~kg} \) of coke is
(Assume coke to be \( 100 \% \) carbon.)
(Given enthalpies of combustion of \( \mathrm{CO}_{2}, \mathrm{CO} \) and \( \mathrm{H}_{2} \) as \( 393.5 \mathrm{~kJ}, 285 \mathrm{~kJ}, 285 \mathrm{~kJ} \) respectively
all at \( 298 \) K.)

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