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A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011]
- a)1.8 atm
- b)3 atm
- c)0.3 atm
- d)0.18 atm
Correct answer is option 'A'. Can you explain this answer?
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A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of th...
Kp depends upon the partial pressure of reactants and products so first calculate their partial pressure and then, calculate Kp,
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A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of th...
Given:
- Temperature (T) = 1000 K
- Initial pressure of CO2 (P1) = 0.5 atm
- Final pressure at equilibrium (P) = 0.8 atm
To find:
- The value of Kp
Step 1: Write the balanced chemical equation
CO2(g) + C(s) ⇌ 2CO(g)
Step 2: Write the expression for the equilibrium constant (Kp)
Kp = (PCO)^2 / (PCO2 * Pgraphite)
Step 3: Determine the partial pressures of CO and CO2 at equilibrium
Let x be the number of moles of CO formed.
Since 1 mole of CO2 reacts to form 2 moles of CO, the moles of CO2 remaining will be (0.5 - x) and the moles of CO will be 2x.
The partial pressure of CO (PCO) can be calculated using the ideal gas law:
PCO = (nCO * RT) / V
where nCO is the number of moles of CO, R is the ideal gas constant, T is the temperature, and V is the volume.
Similarly, the partial pressure of CO2 (PCO2) can be calculated:
PCO2 = (nCO2 * RT) / V
where nCO2 is the number of moles of CO2.
Step 4: Substitute the partial pressures into the equilibrium constant expression
Kp = [(2x / V)^2] / [(0.5 - x) / V * Pgraphite]
Simplifying,
Kp = 4x^2 / [(0.5 - x) * Pgraphite]
Step 5: Calculate the value of x
To find the value of x, we can use the fact that the total pressure at equilibrium (P) is 0.8 atm.
According to Dalton's law of partial pressures, the sum of the partial pressures of all the gases in the mixture is equal to the total pressure.
P = PCO + PCO2 = (2x / V) + [(0.5 - x) / V * Pgraphite]
0.8 = (2x / V) + [(0.5 - x) / V * Pgraphite]
0.8V = 2x + (0.5 - x) * Pgraphite
0.8V = 2x + 0.5Pgraphite - xPgraphite
0.8V - 0.5Pgraphite = x(2 - Pgraphite)
Step 6: Substitute the values into the equilibrium constant expression
Kp = 4x^2 / [(0.5 - x) * Pgraphite]
Kp = 4(0.8V - 0.5Pgraphite)^2 / [(0.5 - (0.8V - 0.5Pgraphite)) * Pgraphite]
Kp = 4(0.8V - 0.5Pgraphite)^2 / [(0.5 - 0.8V + 0.5Pgraphite) * Pgraphite]
Kp = 4(0.8V - 0.5Pgraphite)^2 / [0.5Pgraphite - 0.8VP
- Temperature (T) = 1000 K
- Initial pressure of CO2 (P1) = 0.5 atm
- Final pressure at equilibrium (P) = 0.8 atm
To find:
- The value of Kp
Step 1: Write the balanced chemical equation
CO2(g) + C(s) ⇌ 2CO(g)
Step 2: Write the expression for the equilibrium constant (Kp)
Kp = (PCO)^2 / (PCO2 * Pgraphite)
Step 3: Determine the partial pressures of CO and CO2 at equilibrium
Let x be the number of moles of CO formed.
Since 1 mole of CO2 reacts to form 2 moles of CO, the moles of CO2 remaining will be (0.5 - x) and the moles of CO will be 2x.
The partial pressure of CO (PCO) can be calculated using the ideal gas law:
PCO = (nCO * RT) / V
where nCO is the number of moles of CO, R is the ideal gas constant, T is the temperature, and V is the volume.
Similarly, the partial pressure of CO2 (PCO2) can be calculated:
PCO2 = (nCO2 * RT) / V
where nCO2 is the number of moles of CO2.
Step 4: Substitute the partial pressures into the equilibrium constant expression
Kp = [(2x / V)^2] / [(0.5 - x) / V * Pgraphite]
Simplifying,
Kp = 4x^2 / [(0.5 - x) * Pgraphite]
Step 5: Calculate the value of x
To find the value of x, we can use the fact that the total pressure at equilibrium (P) is 0.8 atm.
According to Dalton's law of partial pressures, the sum of the partial pressures of all the gases in the mixture is equal to the total pressure.
P = PCO + PCO2 = (2x / V) + [(0.5 - x) / V * Pgraphite]
0.8 = (2x / V) + [(0.5 - x) / V * Pgraphite]
0.8V = 2x + (0.5 - x) * Pgraphite
0.8V = 2x + 0.5Pgraphite - xPgraphite
0.8V - 0.5Pgraphite = x(2 - Pgraphite)
Step 6: Substitute the values into the equilibrium constant expression
Kp = 4x^2 / [(0.5 - x) * Pgraphite]
Kp = 4(0.8V - 0.5Pgraphite)^2 / [(0.5 - (0.8V - 0.5Pgraphite)) * Pgraphite]
Kp = 4(0.8V - 0.5Pgraphite)^2 / [(0.5 - 0.8V + 0.5Pgraphite) * Pgraphite]
Kp = 4(0.8V - 0.5Pgraphite)^2 / [0.5Pgraphite - 0.8VP
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A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer?
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A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer?.
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of Kp is: [AIEEE 2011] a)1.8 atmb)3 atmc)0.3 atmd)0.18 atmCorrect answer is option 'A'. Can you explain this answer?.
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