Question

When $20\, g$ of $CaCO _{3}$ were put into $10\, L$ flask and heated to $800^{\circ} C , 35 \%$ of $CaCO _{3}$ remained unreacted at equilibrium, $K_{P}$ for decomposition of $CaCO _{3}$ is

• A. 1.145 atm
• B. 0.145 atm
• C. 2.146 atm
• D. 3.145 atm

Answer 1

Answer: (A)

$\therefore $ Moles of $CO _{2}$ formed $=\frac{20 \times 65}{10^{4}}=1.3 \times 10^{-1}$

$\because P V=n R T$

$P_{ CO } =\frac{1.3 \times 10^{-1}}{10} \times 0.0821 \times 1073=1.145$ atm

Now $K_{p}=P_{ CO }=1.145$ atm