2. Solid CaCO3 upon heating decomposes to solid CaO and CO₂ gas. If the pressure &molot sealed container was found to be 0.76 atm at equilibrium at a temperature equilibrium constant for this reaction. Calculate both Kp and Kc. 1.6 in a 1 L of 560 K what is the BSP=
2. Solid CaCO3 upon heating decomposes to solid CaO and CO₂ gas. If the pressure &molot sealed container was found to be 0.76 atm at equilibrium at a temperature equilibrium constant for this reaction. Calculate both Kp and Kc. 1.6 in a 1 L of 560 K what is the BSP=
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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Transcribed Image Text:**Problem Statement:**
Solid CaCO₃, upon heating, decomposes to solid CaO and CO₂ gas. If the pressure in a 1L sealed container was found to be 0.76 atm at equilibrium at a temperature of 560 K, what is the equilibrium constant for this reaction? Calculate both Kp and Kc.
**Notes:**
- The reaction involves solid calcium carbonate decomposing into calcium oxide and carbon dioxide gas.
- The pressure of carbon dioxide at equilibrium is given as 0.76 atm.
- The container volume is 1 liter, and the temperature is 560 K.
**Solution Approach:**
- Since solid substances do not appear in the equilibrium expression, focus only on the gaseous component, CO₂.
- Use the provided pressure to determine Kp.
- Relate Kp to Kc using the ideal gas constant and temperature for conversion between the two expressions.
**Calculations:**
- Kp = P(CO₂) since the reaction produces one mole of gas from a solid, making it straightforward.
- For conversion between Kp and Kc, use the formula: Kp = Kc(RT)^Δn, where Δn is the change in moles of gas (here Δn = 1), R is the ideal gas constant, and T is temperature in Kelvin.
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