Question

$20.0$grams of $CaC{O}_3\left(s\right)$ were placed in a closed vessel, heated & maintained at ${727}^o$C under equilibrium $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$ and it is found that $75\%$ of $CaC{O}_3$ was decomposed. What is the value of $K_p$? The volume of the container was $15$ litre.

Answer 1

Molar mass of $Ca{CO}_3=100g$ ; No. of moles taken $=\frac{20}{100}=\frac{1}{5}=0.2$

The given reaction is :-

$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right)$

Initial moles : $0.2$ $0$ $0$

At eqm : $(0.2-x)$ $x$ $x$

$=0.5$ $1.5$ $1.5$

Now, given that $75$% of ${CaCO}_3$ is decomposed.

$\Rightarrow x=0.75of0.2=1.5$

So, At eqm moles of ${CaCO}_3$ left $=0.2-1.5=0.5$

Now, $T={727}^{0}C=727+273=1000K$

Now, $K_P=p_{{CO}_2}=\frac{nRT}{v}$ (by ideal gas equation)

$=\frac{1.5\times 0.0821\times 1000}{15}=8.21$