**Equilibrium Problem: Decomposition of Calcium Carbonate****Chemical Equation:**\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]**Initial Conditions:**- 0.100 mol of \(\text{CaCO}_3\) and 0.100 mol of \(\text{CaO}\) are placed in a 10.0 L evacuated container.- The container is heated to 385 K.- At equilibrium, the pressure of \(\text{CO}_2\) is 0.220 atm.**Disturbance:**- 0.270 atm of \(\text{CO}_2\) is added while maintaining a constant temperature.**Question:**- What will be the final mass of \(\text{CaCO}_3\) once the system reaches equilibrium again?**Answer Choices:**- 5.00 g (Your answer)- 15.01 g- 8.42 g- 11.59 g (Correct answer)- 10.01 g**Explanation:**This problem illustrates how changes in pressure can affect chemical equilibrium. When additional \(\text{CO}_2\) is introduced into the system, the equilibrium will shift according to Le Chatelier's Principle to counteract the change. The correct answer indicates the new mass of \(\text{CaCO}_3\) once equilibrium is reestablished.

CaCO3(s) CaO(s) + CO₂(g) 0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.270 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3?

CaCO3(s) CaO(s) + CO₂(g) 0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.270 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

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**Equilibrium Problem: Decomposition of Calcium Carbonate**

**Chemical Equation:**
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

**Initial Conditions:**
- 0.100 mol of \(\text{CaCO}_3\) and 0.100 mol of \(\text{CaO}\) are placed in a 10.0 L evacuated container.
- The container is heated to 385 K.
- At equilibrium, the pressure of \(\text{CO}_2\) is 0.220 atm.

**Disturbance:**
- 0.270 atm of \(\text{CO}_2\) is added while maintaining a constant temperature.

**Question:**
- What will be the final mass of \(\text{CaCO}_3\) once the system reaches equilibrium again?

**Answer Choices:**
- 5.00 g (Your answer)
- 15.01 g
- 8.42 g
- 11.59 g (Correct answer)
- 10.01 g

**Explanation:**
This problem illustrates how changes in pressure can affect chemical equilibrium. When additional \(\text{CO}_2\) is introduced into the system, the equilibrium will shift according to Le Chatelier's Principle to counteract the change. The correct answer indicates the new mass of \(\text{CaCO}_3\) once equilibrium is reestablished.

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Step 1: Ideal gas equation

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Step 2: Calculation for the moles of carbon dioxide formed in the first case

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Step 3: Calculation for the moles of Calcium carbonate remains after the equilibrium reached in first case

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Step 4: Calculation for the value of equilibrium constant

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Step 5: Calculation for the pressure of carbon dioxide that has reacted back

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Step 6: Calculation for the moles of Carbon dioxide those have reacted back

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Step 7: Calculation for the moles of Calcium carbonate after the reestablishing of equilibrium

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Step 8: Calculation for the mass of Calcium carbonate after the reestablishing of equilibrium

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