Question

$H_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to H_2{O}_{\left(g\right)}$

$\Delta H=241.8kJ$

$C{O}_{\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to C{O}_{2\left(g\right)}$

$\Delta H=283kJ$

The heat evolved during the combustion of $112$ litre of water gas (mixture of equal volume of $H_2$ and $CO$) is:

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

Answer 1

Answer: (C)

$1312kJ$

Solution:- (C) $1312KJ$

Molecular formula of water gas $=H_2+CO$

Combustion of $H_2$-

$H_2+\frac{1}{2}{O}_2⟶H_{2}O;\Delta H=241.3KJ$

Combustion of $CO$-

$CO+\frac{1}{2}{O}_2⟶C{O}_2;\Delta H=283KJ$

$1$ mole of $H_2$ and $CO$ each gives,

$\Delta H=283+\left(241.8\right)=524.8KJ$

$1$ mole of $H_2=22.4L$

Similarly,

$1$ mole of $CO=22.4L$

$\therefore$ Total volume $=22.4+22.4=44.8L$

Therefore,

Heat evolved during combustion of $44.8L$ of water gas $=524.8KJ$

Heat evolved during combustion of $112L$ of water gas $=\frac{524.8\times 112}{44.8}=1312KJ$

Hence the heat evolved during the combustion of $112$ litre of water gas at STP is $1312KJ$.