Question

When $20g$ of ${CaCO}_3$ were put into $10$ litre flask and heated to $800$, $30$% of ${CaCO}_3$ remained unreacted at equilibrium. $K_p$ for decomposition of ${CaCO}_3$ will be:

• A. $1.145atm$
• B. $1.231atm$
• C. $2.146atm$
• D. $3.145atm$

Answer 1

Answer: (B)

$1.231atm$

${CaCO}_{3_{\left(s\right)}}\rightleftharpoons {CaO}_{\left(s\right)}+{CO}_{2_{\left(g\right)}}$

No. of moles of ${CaCO}_3=\frac{20}{100}=0.2$ moles

At equilibrium, $30$% of ${CaCO}_3$ unreacted

$\therefore \frac{30}{100}\times 0.2=0.06$ moles

Now, ${CaCO}_{3_{\left(s\right)}}\rightleftharpoons {CaO}_{\left(s\right)}+{CO}_{2_{\left(g\right)}}$

Initial $\to \begin{array}{ccc}0.2& 0& 0\\ 0.06& 0& 0\\ 0.14& 0.14& 0.14\end{array}$

$K_C=0.14$ moles/litre

Now, $K_P=K_C\frac{{\left(RT\right)}^{△ng}}{V}$

$\therefore {}^{}=0.14\times {\left(\frac{0.0821\times 1073}{10}\right)}^1$

$\therefore K_P=1.231atm$