Question

From the following data of $\Delta H$, of the following reactions,

$C\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to CO\left(g\right)AH=-110$ kJ
$C\left(s\right)+H_{2}O\left(g\right)\to CO\left(g\right)+H_2\left(g\right)\Delta H=132$ kJ

What is the mole composition of the mixture of steam and oxygen on being passed over a coke at $1273$ K, keeping the temperature constant?

• A. $0.5:1$
• B. $0.6:1$
• C. $0.8:1$
• D. $1:1$

Answer 1

The correct option is B $0.6:1$

$C\left(s\right)+1/2O_2\left(g\right)\to CO\left(g\right)$ ; ${\Delta H}_1=-100kJ$

$C\left(s\right)+H_{2}O\left(g\right)\to CO\left(g\right)+H_2\left(g\right)$ ; ${\Delta H}_2=132kJ$

So the 1st reaction is exothermic and 2nd reaction is endothermic. So if a mixture of steam and $O_2$ is passed over coke and $T$ is keeping constant, conversion of each of $CO$ should not show ant heat change i.e. heat evolved in reqn $I=$ heat absorbed in reaction

$\therefore$ $n_1\times 2\times 100=n_2\times 132\times 1$

$\therefore$ $\frac{n_1}{n_2}=\frac{132}{110\times 2}\Rightarrow n_1=0.6n_2$

$\therefore n_1:n_2=0.6:1$