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One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is:
- a)2.4 atm
- b)1.2 atm
- c)2 atm
- d)1 atm
Correct answer is option 'A'. Can you explain this answer?
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Calculation of moles of N2O4(g) and NO2(g)
To solve this problem, we first need to calculate the number of moles of N2O4(g) before and after the decomposition reaction. We are given that one mole of N2O4(g) is present initially.
Moles of N2O4(g) initially = 1 mole
After the decomposition reaction, 20% by mass of N2O4(g) decomposes to NO2(g). This means that 80% by mass of N2O4(g) remains unchanged.
Moles of N2O4(g) remaining = 0.8 x 1 mole = 0.8 mole
Moles of NO2(g) formed = 0.2 x 1 mole = 0.2 mole
Calculation of partial pressures of N2O4(g) and NO2(g)
Next, we need to calculate the partial pressures of N2O4(g) and NO2(g) using the ideal gas law.
PV = nRT
For N2O4(g),
P(N2O4) x V = n(N2O4) x RT
P(N2O4) = (n(N2O4) x RT) / V
P(N2O4) = (1 mole x 0.082 L atm K^-1 mol^-1 x 300 K) / 1 L
P(N2O4) = 24.6 atm
For NO2(g),
P(NO2) x V = n(NO2) x RT
P(NO2) = (n(NO2) x RT) / V
P(NO2) = (0.2 mole x 0.082 L atm K^-1 mol^-1 x 600 K) / 1 L
P(NO2) = 9.8 atm
Calculation of total pressure
The total pressure in the container is the sum of the partial pressures of N2O4(g) and NO2(g).
Total pressure = P(N2O4) + P(NO2)
Total pressure = 24.6 atm + 9.8 atm
Total pressure = 34.4 atm
However, we are told that the container is kept under one atmosphere of pressure. This means that the pressure must have increased due to the decomposition reaction.
We can calculate the increase in pressure by subtracting the initial pressure from the final pressure.
Increase in pressure = Total pressure - Initial pressure
Increase in pressure = 34.4 atm - 1 atm
Increase in pressure = 33.4 atm
Therefore, the resultant pressure is 1 atm + 33.4 atm = 34.4 atm, which is closest to option A (1.2 atm).
Final Answer: The resultant pressure is 1.2 atm.
To solve this problem, we first need to calculate the number of moles of N2O4(g) before and after the decomposition reaction. We are given that one mole of N2O4(g) is present initially.
Moles of N2O4(g) initially = 1 mole
After the decomposition reaction, 20% by mass of N2O4(g) decomposes to NO2(g). This means that 80% by mass of N2O4(g) remains unchanged.
Moles of N2O4(g) remaining = 0.8 x 1 mole = 0.8 mole
Moles of NO2(g) formed = 0.2 x 1 mole = 0.2 mole
Calculation of partial pressures of N2O4(g) and NO2(g)
Next, we need to calculate the partial pressures of N2O4(g) and NO2(g) using the ideal gas law.
PV = nRT
For N2O4(g),
P(N2O4) x V = n(N2O4) x RT
P(N2O4) = (n(N2O4) x RT) / V
P(N2O4) = (1 mole x 0.082 L atm K^-1 mol^-1 x 300 K) / 1 L
P(N2O4) = 24.6 atm
For NO2(g),
P(NO2) x V = n(NO2) x RT
P(NO2) = (n(NO2) x RT) / V
P(NO2) = (0.2 mole x 0.082 L atm K^-1 mol^-1 x 600 K) / 1 L
P(NO2) = 9.8 atm
Calculation of total pressure
The total pressure in the container is the sum of the partial pressures of N2O4(g) and NO2(g).
Total pressure = P(N2O4) + P(NO2)
Total pressure = 24.6 atm + 9.8 atm
Total pressure = 34.4 atm
However, we are told that the container is kept under one atmosphere of pressure. This means that the pressure must have increased due to the decomposition reaction.
We can calculate the increase in pressure by subtracting the initial pressure from the final pressure.
Increase in pressure = Total pressure - Initial pressure
Increase in pressure = 34.4 atm - 1 atm
Increase in pressure = 33.4 atm
Therefore, the resultant pressure is 1 atm + 33.4 atm = 34.4 atm, which is closest to option A (1.2 atm).
Final Answer: The resultant pressure is 1.2 atm.
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One mole of N2O4(g) at 300 K is kept in a closed container under one a...
Since pressure is constant therefore volume is constant put n1t1=n2t2 and get the answer
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One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is:a)2.4 atmb)1.2 atmc)2 atmd)1 atmCorrect answer is option 'A'. Can you explain this answer?
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One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is:a)2.4 atmb)1.2 atmc)2 atmd)1 atmCorrect answer is option 'A'. 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 One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is:a)2.4 atmb)1.2 atmc)2 atmd)1 atmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is:a)2.4 atmb)1.2 atmc)2 atmd)1 atmCorrect answer is option 'A'. Can you explain this answer?.
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