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A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.
Correct answer is '6.25'. Can you explain this answer?
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A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L...
H2 + CO2 - mixture of non-reacting gases
2g H2 = 1 mol
176g CO2 = 4 mol
Total no. of moles = 5
Total volume = 2+2 = 4L
Temperature = 27+273 = 300K
Total pressure, P=nRT/V=(5*0.083*300/4)= 31.125 bar
According to Dalton's law, p1 = P*(x1)
Mole fraction of H2=1/(1+4)=1/5
Therefore, p(H2)= 31.125*0.2 = 6.225 bar
2g H2 = 1 mol
176g CO2 = 4 mol
Total no. of moles = 5
Total volume = 2+2 = 4L
Temperature = 27+273 = 300K
Total pressure, P=nRT/V=(5*0.083*300/4)= 31.125 bar
According to Dalton's law, p1 = P*(x1)
Mole fraction of H2=1/(1+4)=1/5
Therefore, p(H2)= 31.125*0.2 = 6.225 bar
Most Upvoted Answer
A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L...
Solution:
Given:
Volume of vessel 1 (V1) = 2 L
Volume of vessel 2 (V2) = 2 L
Amount of H2 gas in vessel 1 (n1) = 2 g
Amount of CO2 gas in vessel 2 (n2) = 176 g
Temperature (T) = 27°C or 300K (since temperature is in Celsius, it needs to be converted to Kelvin)
To find:
Partial pressure of H2 at equilibrium
Step 1: Calculate the number of moles of H2 and CO2
We know that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
For H2 gas in vessel 1:
P1V1 = n1RT
n1 = (P1V1)/(RT)
For CO2 gas in vessel 2:
P2V2 = n2RT
n2 = (P2V2)/(RT)
Substituting the given values, we get:
n1 = (P1×2)/(0.0821×300)
n1 = 0.081P1
n2 = (P2×2)/(0.0821×300)
n2 = 0.081P2
Step 2: Write the balanced chemical equation
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
Step 3: Calculate the equilibrium constant (Kp)
Kp = (P_H2O × P_CO) / (P_H2 × P_CO2)
Since the volume of both vessels is the same, we can assume that the total pressure (P_total) remains constant. Therefore, we can write:
P_total = P_H2 + P_CO2 + P_H2O + P_CO
Also, at equilibrium, the moles of H2, CO2, H2O, and CO are related by the stoichiometric coefficients in the balanced equation:
n_H2O = n_CO
n_H2 = n1 - n_H2O
n_CO2 = n2 - n_CO
Substituting the expressions for n1, n2, n_H2O, n_CO, n_H2, and n_CO2 in terms of the partial pressures, we get:
Kp = (P_H2O × P_CO) / (P_H2 × P_CO2)
Kp = [(P_total - P_H2 - P_CO2) × P_CO] / [(0.081P1 - P_CO) × (0.081P2 - P_H2)]
Step 4: Calculate the equilibrium partial pressure of H2 (P_H2)
We can solve the quadratic equation obtained by substituting the given values and simplifying:
Kp = (P_H2O × P_CO) / (P_H2 × P_CO2)
6.375 × 10^-4 = [(P_total - P_H2 - 1.110) × 1.110] / [(0.081P1 - 1.110) × (2.184 - P_H2)]
6.375 × 10^-4 = [(P_total - P_H2 - 1.110) × 1.110] / [0.
Given:
Volume of vessel 1 (V1) = 2 L
Volume of vessel 2 (V2) = 2 L
Amount of H2 gas in vessel 1 (n1) = 2 g
Amount of CO2 gas in vessel 2 (n2) = 176 g
Temperature (T) = 27°C or 300K (since temperature is in Celsius, it needs to be converted to Kelvin)
To find:
Partial pressure of H2 at equilibrium
Step 1: Calculate the number of moles of H2 and CO2
We know that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
For H2 gas in vessel 1:
P1V1 = n1RT
n1 = (P1V1)/(RT)
For CO2 gas in vessel 2:
P2V2 = n2RT
n2 = (P2V2)/(RT)
Substituting the given values, we get:
n1 = (P1×2)/(0.0821×300)
n1 = 0.081P1
n2 = (P2×2)/(0.0821×300)
n2 = 0.081P2
Step 2: Write the balanced chemical equation
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
Step 3: Calculate the equilibrium constant (Kp)
Kp = (P_H2O × P_CO) / (P_H2 × P_CO2)
Since the volume of both vessels is the same, we can assume that the total pressure (P_total) remains constant. Therefore, we can write:
P_total = P_H2 + P_CO2 + P_H2O + P_CO
Also, at equilibrium, the moles of H2, CO2, H2O, and CO are related by the stoichiometric coefficients in the balanced equation:
n_H2O = n_CO
n_H2 = n1 - n_H2O
n_CO2 = n2 - n_CO
Substituting the expressions for n1, n2, n_H2O, n_CO, n_H2, and n_CO2 in terms of the partial pressures, we get:
Kp = (P_H2O × P_CO) / (P_H2 × P_CO2)
Kp = [(P_total - P_H2 - P_CO2) × P_CO] / [(0.081P1 - P_CO) × (0.081P2 - P_H2)]
Step 4: Calculate the equilibrium partial pressure of H2 (P_H2)
We can solve the quadratic equation obtained by substituting the given values and simplifying:
Kp = (P_H2O × P_CO) / (P_H2 × P_CO2)
6.375 × 10^-4 = [(P_total - P_H2 - 1.110) × 1.110] / [(0.081P1 - 1.110) × (2.184 - P_H2)]
6.375 × 10^-4 = [(P_total - P_H2 - 1.110) × 1.110] / [0.
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Community Answer
A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L...
H2 + CO2 - mixture of non-reacting gases
2g H2 = 1 mol
176g CO2 = 4 mol
Total no. of moles = 5
Total volume = 2+2 = 4L
Temperature = 27+273 = 300K
Total pressure, P=nRT/V=(5*0.083*300/4)= 31.125 bar
According to Dalton's law, p1 = P*(x1)
Mole fraction of H2=1/(1+4)=1/5
Therefore, p(H2)= 31.125*0.2 = 6.225 bar
2g H2 = 1 mol
176g CO2 = 4 mol
Total no. of moles = 5
Total volume = 2+2 = 4L
Temperature = 27+273 = 300K
Total pressure, P=nRT/V=(5*0.083*300/4)= 31.125 bar
According to Dalton's law, p1 = P*(x1)
Mole fraction of H2=1/(1+4)=1/5
Therefore, p(H2)= 31.125*0.2 = 6.225 bar
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A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer?
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A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer?.
A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer?.
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A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer?, a detailed solution for A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer? has been provided alongside types of A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.Correct answer is '6.25'. Can you explain this answer? theory, EduRev gives you an
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