Question

For the equilibrium :
$CaC{O}_{3_{(}s)}\leftrightharpoons Ca{O}_{(s)}+C{O}_{2(g)}$;
$K_p=1.64atmat1000K$
$50g$ of $CaC{O}_3$ in a $10litre$ closed vessel is heated to $1000K$. Percentage of $CaC{O}_3$ that remains unreacted at equilibrium is (Given $R=0.082Latm{K}^{-1}mo{l}^{-1}$)

• A. 50
• B. 20
• C. 40
• D. 60

Answer 1

Answer: (D)

$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$

$K_p=p_{C{O}_2}=\frac{n}{V}RT$

$\therefore 1.64=\frac{n}{10}\times 0.082\times 1000$

$\therefore n=\frac{1.64\times 10}{0.082\times 1000}=0.2$

i.e., Number of moles of $C{O}_2=0.2$

$CaC{O}_3\rightleftharpoons CaO+C{O}_2$

Initially $50g=\frac{50}{100}=0.5mol0$

At equilibrium $0.2mol0.2$

$\therefore$ Unreacted $CaC{O}_3=0.5-0.2=0.3mol$

and $\%$ of unreacted $CaC{O}_3=\frac{0.3}{0.5}\times 100$

$=60\%$