Question

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 all at $298\, K$.)

• A. $0.69 : 1$
• B. $0.96 : 1$
• C. $0.79 : 1$
• D. $0.86 : 1$

Answer 1

Answer: (A)

Number of moles in $1 kg$ coke

$=\frac{1000}{12}=83.33\, mol$

(i) For the combustion of $1 kg$ of coke

$C + O _{2} \longrightarrow CO _{2} ;\, \Delta H=393.5\, kJ$

$\Rightarrow $ heat liberated from 1 mole coke

$=393.5\, kJ$

$\therefore $ heat liberated from $83.33$ mole coke

$=(393.5 \times 83.33)\, kJ$

(ii) For the burning of water gas

$C + H _{2} O \longrightarrow \underbrace{ CO + H _{2}}_{\text {Water gas }}$

$CO + H _{2}+ O _{2} \longrightarrow CO _{2}+ H _{2} O$

$\Delta H=285+285=570\, kJ$

$\therefore $ Ratio of heat liberated from burning of water gas obtained from $1\, kg$ of coke

$H_{2}=(570 \times 83.33)\, kJ$

$\therefore $ Required ratio $= H _{1}: H _{2}$

$=393.5: 570$

$=0.69: 1$