When 20 g of CaCO 3 were put into 10 L flask and heated to 800° C , 35 % of CaCO 3 remained unreacted at equilibrium, KP for decomposition of CaCO 3 is

Question

When 20 g of CaCO 3 were put into 10 L flask and heated to 800° C , 35 % of CaCO 3 remained unreacted at equilibrium, KP for decomposition of CaCO 3 is (A) 1.145 atm (B) 0.145 atm (C) 2.146 atm (D) 3.145 atm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/24d41e7c49511f0572f85ab5769c4ca6-.png" /> ∴ Moles of CO 2 formed =(20 × 65/104)=1.3 × 10-1 ∵ P V=n R T P CO =(1.3 × 10-1/10) × 0.0821 × 1073=1.145 atm Now Kp=P CO =1.145 atm

Q. When $20 g$ of $C a C O_{3}$ were put into $10 L$ flask and heated to $80 0^{\circ} C, 35%$ of $C a C O_{3}$ remained unreacted at equilibrium, $K_{P}$ for decomposition of $C a C O_{3}$ is

1.145 atm

0.145 atm

2.146 atm

3.145 atm

$∴$ Moles of $C O_{2}$ formed $= \frac{20 \times 65}{1 0 ^{4}} = 1.3 \times 1 0^{- 1}$

$∵ P V = n RT$

$P_{CO} = \frac{1.3 \times 1 0 ^{- 1}}{10} \times 0.0821 \times 1073 = 1.145$ atm

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

Q. When 20g of CaCO3​ were put into 10L flask and heated to 800∘C,35% of CaCO3​ remained unreacted at equilibrium, KP​ for decomposition of CaCO3​ is