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At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,
At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N2O4 (g) is thus,
- a)25%
- b)50%
- c)66.66%
- d)33.33%
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At e...
Let the initial volume of N2O4 be x and initial volume of NO2 is 0
If the degree of dissociation is a, then the final volume of N2O4 is x(1−a) and NO2 is 2ax.
Initial
It equilibrium
N2O4 ⟶ 2NO2
x 0
x(1−a) 2ax
Total initial volume =x+0=x
Final volume =x(1−a)+2ax=x+ax=x(1+a)
It is given that the initial volume is 25% less than the final volume
x=0.75×(1+a)
1+a=1.33
a=0.33
So %age dissociation = 33.33%
If the degree of dissociation is a, then the final volume of N2O4 is x(1−a) and NO2 is 2ax.
Initial
It equilibrium
N2O4 ⟶ 2NO2
x 0
x(1−a) 2ax
Total initial volume =x+0=x
Final volume =x(1−a)+2ax=x+ax=x(1+a)
It is given that the initial volume is 25% less than the final volume
x=0.75×(1+a)
1+a=1.33
a=0.33
So %age dissociation = 33.33%
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At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At e...
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At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At e...
Given:
- Temperature (T) = 273 K
- Pressure (P) = 1 atm
- Volume of N2O4 (g) = 1 L
Equation:
N2O4(g) ⇌ 2NO2(g)
Explanation:
At equilibrium, the volume of the original gas is 25% less than the existing volume. To determine the percentage decomposition of N2O4, we need to calculate the change in volume.
Let's assume the initial volume of N2O4 is V liters. According to the given information, the original volume of N2O4 at equilibrium is 25% less than V, which means the new volume is 0.75V liters.
Since the stoichiometric coefficient of N2O4 is 1 and the stoichiometric coefficient of NO2 is 2 in the balanced equation, the number of moles of N2O4 (g) will be equal to the number of moles of NO2(g) formed.
Step 1: Calculate the number of moles of N2O4:
Using the ideal gas equation: PV = nRT
Where P = pressure, V = volume, n = number of moles, R = ideal gas constant, and T = temperature.
For N2O4(g):
PV = nN2O4RT
1 atm * V L = nN2O4 * 0.0821 L.atm/mol.K * 273 K
V = nN2O4 * 22.414 L/mol
Therefore, the number of moles of N2O4 (g) is equal to the volume of N2O4 (g) in liters.
Step 2: Calculate the number of moles of NO2:
Since the stoichiometric coefficient of N2O4 is 1 and the stoichiometric coefficient of NO2 is 2, the number of moles of NO2 formed will be twice the number of moles of N2O4 decomposed.
Number of moles of NO2 = 2 * (0.25V) = 0.5V
Step 3: Calculate the percentage decomposition of N2O4:
Percentage decomposition = (moles of N2O4 decomposed / initial moles of N2O4) * 100
Percentage decomposition = (0.25V / V) * 100
Percentage decomposition = 25%
Therefore, the correct answer is option 'D' - 33.33%.
- Temperature (T) = 273 K
- Pressure (P) = 1 atm
- Volume of N2O4 (g) = 1 L
Equation:
N2O4(g) ⇌ 2NO2(g)
Explanation:
At equilibrium, the volume of the original gas is 25% less than the existing volume. To determine the percentage decomposition of N2O4, we need to calculate the change in volume.
Let's assume the initial volume of N2O4 is V liters. According to the given information, the original volume of N2O4 at equilibrium is 25% less than V, which means the new volume is 0.75V liters.
Since the stoichiometric coefficient of N2O4 is 1 and the stoichiometric coefficient of NO2 is 2 in the balanced equation, the number of moles of N2O4 (g) will be equal to the number of moles of NO2(g) formed.
Step 1: Calculate the number of moles of N2O4:
Using the ideal gas equation: PV = nRT
Where P = pressure, V = volume, n = number of moles, R = ideal gas constant, and T = temperature.
For N2O4(g):
PV = nN2O4RT
1 atm * V L = nN2O4 * 0.0821 L.atm/mol.K * 273 K
V = nN2O4 * 22.414 L/mol
Therefore, the number of moles of N2O4 (g) is equal to the volume of N2O4 (g) in liters.
Step 2: Calculate the number of moles of NO2:
Since the stoichiometric coefficient of N2O4 is 1 and the stoichiometric coefficient of NO2 is 2, the number of moles of NO2 formed will be twice the number of moles of N2O4 decomposed.
Number of moles of NO2 = 2 * (0.25V) = 0.5V
Step 3: Calculate the percentage decomposition of N2O4:
Percentage decomposition = (moles of N2O4 decomposed / initial moles of N2O4) * 100
Percentage decomposition = (0.25V / V) * 100
Percentage decomposition = 25%
Therefore, the correct answer is option 'D' - 33.33%.
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At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N2O4 (g) is thus,a)25%b)50%c)66.66%d)33.33%Correct answer is option 'D'. Can you explain this answer?
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At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N2O4 (g) is thus,a)25%b)50%c)66.66%d)33.33%Correct answer is option 'D'. 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 At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N2O4 (g) is thus,a)25%b)50%c)66.66%d)33.33%Correct answer is option 'D'. 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 At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N2O4 (g) is thus,a)25%b)50%c)66.66%d)33.33%Correct answer is option 'D'. Can you explain this answer?.
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