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One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4 (g) decomposes to NO2(g). The resultant pressure is :
- a)1.2 atm
- b)2.4 atm
- c)2.0 atm
- d)1.0 atm
Correct answer is option 'B'. Can you explain this answer?
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One mole of N2O4(g) at 300 K is left in a closed container under one a...
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One mole of N2O4(g) at 300 K is left in a closed container under one a...
Given:
- 1 mole of N2O4(g) at 300 K
- The container is closed and the pressure is 1 atm
- The temperature is increased to 600 K
- 20% by mass of N2O4(g) decomposes to NO2(g)
To find:
The resultant pressure after the decomposition of N2O4(g)
Solution:
Step 1: Write the balanced chemical equation for the decomposition of N2O4(g) into NO2(g):
N2O4(g) → 2NO2(g)
Step 2: Calculate the initial moles of N2O4:
Given that there is 1 mole of N2O4(g) initially.
Step 3: Calculate the moles of N2O4 decomposed:
Since 20% by mass of N2O4 decomposes, we need to calculate the mass of N2O4 and then convert it to moles.
The molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol
Mass of N2O4 decomposed = 20% of the mass of N2O4 = 0.20 * 92.02 g = 18.40 g
Moles of N2O4 decomposed = Mass of N2O4 decomposed / Molar mass of N2O4 = 18.40 g / 92.02 g/mol = 0.20 mol
Step 4: Calculate the moles of N2O4 remaining:
Moles of N2O4 remaining = Initial moles of N2O4 - Moles of N2O4 decomposed
= 1 mol - 0.20 mol = 0.80 mol
Step 5: Calculate the moles of NO2 produced:
Since the balanced chemical equation shows that 1 mole of N2O4 decomposes to produce 2 moles of NO2, the number of moles of NO2 produced is twice the number of moles of N2O4 decomposed.
Moles of NO2 produced = 2 * Moles of N2O4 decomposed = 2 * 0.20 mol = 0.40 mol
Step 6: Calculate the total moles of gas:
Total moles of gas = Moles of N2O4 remaining + Moles of NO2 produced
= 0.80 mol + 0.40 mol = 1.20 mol
Step 7: Apply the ideal gas law to find the resultant pressure:
PV = nRT
P = (nRT) / V
Given that the initial pressure (P) is 1 atm, the number of moles (n) is 1.20 mol, the gas constant (R) is 0.0821 atm·L/mol·K, and the temperature (T) is 600 K.
P = (1.20 mol * 0.0821 atm·L/mol·K * 600 K) / V
Since the volume (V) is constant, we can write:
P ∝ nT
P2
- 1 mole of N2O4(g) at 300 K
- The container is closed and the pressure is 1 atm
- The temperature is increased to 600 K
- 20% by mass of N2O4(g) decomposes to NO2(g)
To find:
The resultant pressure after the decomposition of N2O4(g)
Solution:
Step 1: Write the balanced chemical equation for the decomposition of N2O4(g) into NO2(g):
N2O4(g) → 2NO2(g)
Step 2: Calculate the initial moles of N2O4:
Given that there is 1 mole of N2O4(g) initially.
Step 3: Calculate the moles of N2O4 decomposed:
Since 20% by mass of N2O4 decomposes, we need to calculate the mass of N2O4 and then convert it to moles.
The molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol
Mass of N2O4 decomposed = 20% of the mass of N2O4 = 0.20 * 92.02 g = 18.40 g
Moles of N2O4 decomposed = Mass of N2O4 decomposed / Molar mass of N2O4 = 18.40 g / 92.02 g/mol = 0.20 mol
Step 4: Calculate the moles of N2O4 remaining:
Moles of N2O4 remaining = Initial moles of N2O4 - Moles of N2O4 decomposed
= 1 mol - 0.20 mol = 0.80 mol
Step 5: Calculate the moles of NO2 produced:
Since the balanced chemical equation shows that 1 mole of N2O4 decomposes to produce 2 moles of NO2, the number of moles of NO2 produced is twice the number of moles of N2O4 decomposed.
Moles of NO2 produced = 2 * Moles of N2O4 decomposed = 2 * 0.20 mol = 0.40 mol
Step 6: Calculate the total moles of gas:
Total moles of gas = Moles of N2O4 remaining + Moles of NO2 produced
= 0.80 mol + 0.40 mol = 1.20 mol
Step 7: Apply the ideal gas law to find the resultant pressure:
PV = nRT
P = (nRT) / V
Given that the initial pressure (P) is 1 atm, the number of moles (n) is 1.20 mol, the gas constant (R) is 0.0821 atm·L/mol·K, and the temperature (T) is 600 K.
P = (1.20 mol * 0.0821 atm·L/mol·K * 600 K) / V
Since the volume (V) is constant, we can write:
P ∝ nT
P2
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One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer?
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One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer?.
One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is :a)1.2 atmb)2.4 atmc)2.0 atmd)1.0 atmCorrect answer is option 'B'. Can you explain this answer?.
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