NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
NTA Abhyas 2020 · Chemistry · Chemical Equilibrium
Equilibrium constant K_ p for the reaction CaCO_ 3 ⇌ CaO + CO_ 2 is...
Equilibrium constant K_{p} for the reaction {\text{CaCO}_{3} ⇌ \text{CaO} + {\text{CO}_{2} is 0.82 atm at If 1 mole of CaCO_{3} is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO_{3} would dissociate at equilibrium?
Options
- A0.2 g
- B80 g
- C20 g
- D50 g
Correct answer
C. 20 g
Step-by-step solution
{\text{CaCO}_{3} ⇌ \text{CaO}\text{+}{\text{CO}_{2}
{K}_{p} = {p}_{{\text{CO}_{2}} = 0.82\text{atm}
0.82\text{atm} \times 20\text{L} = {n}_{{\text{CO}_{2}} \times 0.082 \times 1000
No. of moles CO2 = no. of moles of CaCO3 decomposed = mole
Amount of CaCO3 decomposed
{K}_{p} = {p}_{{\text{CO}_{2}} = 0.82\text{atm}
0.82\text{atm} \times 20\text{L} = {n}_{{\text{CO}_{2}} \times 0.082 \times 1000
No. of moles CO2 = no. of moles of CaCO3 decomposed = mole
Amount of CaCO3 decomposed