Question

From the data of $\Delta H$ of the following reactions
$C\left(s\right)+1/2O_2\left(g\right)\to CO\left(g\right);\Delta H=-110$ kJ
and $C\left(s\right)+H_{2}O\left(g\right)\to CO\left(g\right)+H_2\left(g\right);\Delta H=132$ kJ
Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at
1273 K, keeping the reaction temperature constant.

• A. mole % $O_2\left(g\right)=37.5,H_{2}O\left(g\right)=54.6$
• B. mole % $O_2\left(g\right)=45.5,H_{2}O\left(g\right)=62.5$
• C. mole % $O_2\left(g\right)=37.5,H_{2}O\left(g\right)=62.5$
• D. mole % $O_2\left(g\right)=45.5,H_{2}O\left(g\right)=54.6$

Answer 1

The correct option is C mole % $O_2\left(g\right)=37.5,H_{2}O\left(g\right)=62.5$

The first reaction is exothermic and second is endothermic.If a mixture of steam and
$O_2$ is passed over coke and $T$ is kept constant, conversion of each to $CO$
should not show any heat change i.e.,total heat evolved in $I=$ total heat absorbed in $II$.
$n_1\times 2\times 110=n_2\times 132\times 1$
$\therefore \frac{n_1}{n_2}=\frac{132}{110\times 2}=\frac{0.6}{1}\Rightarrow n_1=0.6n_2$
$n_1=$moles of $O_2$ used during reaction.
$n_2=$moles of $H_{2}O$ used during reaction.
$n_1=0.6n_2$
Try for given option to get $C$ as answer.