Question

At $817$, $K_p$ for the reaction between $C{O}_2\left(g\right)$ and excess hot graphite $\left(s\right)$ is 10 atm.

(a) The sum of equilibrium concentration of the gases at ${817}^{o}C$ is X M and a total pressure of 5 atm.
(b) At Y atm total pressure, the gas contains $5\%$ $C{O}_2$ by volume.

The value of $1000(X+Y)$ is ________.

Answer 1

(a) The given equilibrium is :
$C{O}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right)$
Initial mole $1$ $0$
Final mole $(1-\alpha )$ $2\alpha$

$K_p=\frac{{\left(n_{CO}\right)}^2}{n_{C{O}_2}}\times {\left[\frac{P}{\sum n}\right]}^1=\frac{{\left(2\alpha \right)}^2}{(1-\alpha )}\times \left[\frac{5}{1+\alpha }\right]$

$10=\frac{20{\alpha }^2}{1-{\alpha }^2}$

or $10-10{\alpha }^2=20{\alpha }^2$

$\therefore {\alpha }^2=\frac{10}{30}=\frac{1}{3}$

$\therefore \alpha =\sqrt{\frac{1}{3}}=0.577$

Thus, mole of $C{O}_2$ at equilibrium $=1-\alpha =1-0.577=0.423$ and mole of $CO$ at equilibrium $=2\alpha =2\times 0.577=1.154$

$\therefore$ Total mole of gases at equilibrium $=0.423+1.154=1.577$

Using $PV=nRT$ at equilibrium

$5\times V=1.577\times 0.0821\times 1090$

$\therefore V=28.23litre$

$\therefore \left[C{O}_2\right]=\frac{0.423}{28.23}=0.015M$

$\left[CO\right]=\frac{1.154}{28.23}=0.041M$
(b) At 5% $C{O}_2$ by volume $P_{C{O}_2}^{\prime }=\frac{5}{100}\times P$ (where $P$ is equilibrium pressure)
and $P_{CO}^{\prime }=\frac{95}{100}\times P$
Now, $K_p=\frac{{\left(P_{CO}^{\prime }\right)}^2}{P_{C{O}_2}^{\prime }}=\frac{95\times 95\times P^2\times 100}{100\times 100\times 5\times P}=10$
$\therefore P=0.554atm$
$X=0.041+0.015=0.056M$; $Y=0.554atm$
$1000(X+Y)=1000\left(0.61\right)=610$