Question

The heat evolved during the combustion of $112\, litre$ of water gas (mixture of equal volume of
$H _{2}$ and $CO$ ) is $H _{2(g)}+ O _{2(g)} \rightarrow H _{2} O _{(g)} ; \Delta H=-241.8\, kJ$
$CO _{(g)}+ O _{2(g)} \rightarrow CO _{2(g)} ; \Delta H=-283\, kJ$

• A. $241.8\, kJ$
• B. $283\, kJ$
• C. $-1312\, kJ$
• D. $1586\, kJ$

Answer 1

Answer: (C)

$112$ litre of water gas contains $56$ litre of $H _{2}$ and $56$ litre of $CO$ Heat evolved by combustion of $56$ litre of $H _{2}$

$=\frac{-241.8 \times 56}{22.4}=-604.5\, kJ$

$\Delta H$ for combustion of $56$ litre of

$CO =\frac{-283 \times 56}{22.4}=-707.5 kJ$

Thus $\Delta H$ for combustion of $112$ litre of water gas

$=-604.5+(-707.5)=-1312\, kJ$