Question

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

Given:

$H_2\left(g\right)+1/2O_2\left(g\right)=H_{2}O\left(g\right);\Delta H=-241.8kJ$

$CO\left(g\right)+1/2O_2\left(g\right)=C{O}_2\left(g\right);\Delta H=-283kJ$

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

Answer 1

Answer: (C)

1312 kJ

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,

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

$1$ 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 $1312KJ$.

Hence option C is correct.