Question

From the following data of $\Delta H$, of the following reaction,
$C\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to CO\left(g\right)$ $\Delta H=-110kJ$
$C\left(s\right)+H_{2}O\left(g\right)\to CO\left(g\right)+H_2\left(g\right)$ $\Delta H=132kJ$
What is the mole composition of mixture of steam and oxygen on being passed coke at 1273K, keeping temperature constant?

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

Answer 1

The correct option is B $0.6:1$

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

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

In $1\text{st}$ theaction, heat evolves while in $2^{\text{nd}}$ reaction, heat absosbed.

So if temperature $=1273k$ is to be remained constant, and hence gain or loss by system should be zero.

$\text{Let,}{n}_1\text{moles of steam and}{n}_2\text{moles of}{O}_2\left(g\right)\text{is required.}$

So, Heat evolved $=$ reat absorbed $(\Delta T=0)$

$\begin{array}{cc}\Rightarrow \left(2x110\right)\times n_2=& \left(132\right)\times n_1\\ \therefore \frac{n_1}{n_2}=\frac{2x\parallel 0}{132}& =1.67\end{array}$

[By steichiometry]

But, molar composition of oxygen to steam $=\frac{1}{1.67}=[0.6:1$

So, correct option is (B)