JEE Exam > JEE Questions > PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 ...
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PCl5 dissociates as
PCl5(g) ⇋ PCl3(g) + Cl2(g)
5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)
(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)
PCl5(g) ⇋ PCl3(g) + Cl2(g)
5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)
(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)
Correct answer is '1107'. Can you explain this answer?
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Most Upvoted Answer
PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed ...
We can use the equilibrium expression for Kp:
Kp = (P_PCl3)^1*(P_Cl2)^1 / (P_PCl5)^1
where P represents the partial pressure of each gas at equilibrium, and the superscripts represent the stoichiometric coefficients of each gas in the balanced equation.
We can start by calculating the partial pressures of PCl3 and Cl2 at equilibrium. Since 1 mole of PCl5 dissociates to form 1 mole of PCl3 and 1 mole of Cl2, we can assume that the equilibrium mole fraction of PCl5 is (5-1)/5 = 0.8 and the equilibrium mole fractions of PCl3 and Cl2 are both 0.1 (based on the stoichiometry of the reaction).
Therefore, we can calculate the partial pressures of PCl3 and Cl2 as:
P_PCl3 = 0.1 * 2.46 atm = 0.246 atm
P_Cl2 = 0.1 * 2.46 atm = 0.246 atm
Next, we can calculate the partial pressure of PCl5 at equilibrium by subtracting the partial pressures of PCl3 and Cl2 from the total pressure:
P_PCl5 = 2.46 atm - 0.246 atm - 0.246 atm = 1.968 atm
Finally, we can plug these values into the equilibrium expression for Kp:
Kp = (0.246 atm)^1 * (0.246 atm)^1 / (1.968 atm)^1
Kp = 0.0125 atm
Therefore, the equilibrium constant Kp for the dissociation of PCl5 is 0.0125 atm.
Kp = (P_PCl3)^1*(P_Cl2)^1 / (P_PCl5)^1
where P represents the partial pressure of each gas at equilibrium, and the superscripts represent the stoichiometric coefficients of each gas in the balanced equation.
We can start by calculating the partial pressures of PCl3 and Cl2 at equilibrium. Since 1 mole of PCl5 dissociates to form 1 mole of PCl3 and 1 mole of Cl2, we can assume that the equilibrium mole fraction of PCl5 is (5-1)/5 = 0.8 and the equilibrium mole fractions of PCl3 and Cl2 are both 0.1 (based on the stoichiometry of the reaction).
Therefore, we can calculate the partial pressures of PCl3 and Cl2 as:
P_PCl3 = 0.1 * 2.46 atm = 0.246 atm
P_Cl2 = 0.1 * 2.46 atm = 0.246 atm
Next, we can calculate the partial pressure of PCl5 at equilibrium by subtracting the partial pressures of PCl3 and Cl2 from the total pressure:
P_PCl5 = 2.46 atm - 0.246 atm - 0.246 atm = 1.968 atm
Finally, we can plug these values into the equilibrium expression for Kp:
Kp = (0.246 atm)^1 * (0.246 atm)^1 / (1.968 atm)^1
Kp = 0.0125 atm
Therefore, the equilibrium constant Kp for the dissociation of PCl5 is 0.0125 atm.
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Community Answer
PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed ...
PCl5(g) ⇋ PCl3(g) + Cl2(g)
α = deree of dissociation
Total moles = 5(1 - α) + 5α + α = 5(1 + α)
2 moles of N2 are present, so total moles at equilibrium = 5(1 + α) + 2 ...(1)
By gas equation,
PV = nRT
P = 2.46 atm; V = 200 L
R = 0.082 L atm K-1; T = 600 K
So, n = 10 moles
From equation (1),
5(1 + α) + 2 = 10
Degree of dissociation = α = 3/5 = 0.60
= 1107 × 10-3
α = deree of dissociation
Total moles = 5(1 - α) + 5α + α = 5(1 + α)
2 moles of N2 are present, so total moles at equilibrium = 5(1 + α) + 2 ...(1)
By gas equation,
PV = nRT
P = 2.46 atm; V = 200 L
R = 0.082 L atm K-1; T = 600 K
So, n = 10 moles
From equation (1),
5(1 + α) + 2 = 10
Degree of dissociation = α = 3/5 = 0.60
= 1107 × 10-3
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PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer?
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PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer?.
PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PCl5 dissociates asPCl5(g) PCl3(g) + Cl2(g)5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is ______ × 10-3. (Nearest integer)(Given: R = 0.082 L atm K-1 mol-1; Assume ideal gas behaviour)Correct answer is '1107'. Can you explain this answer?.
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