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Question

# When 0.1 moles of N2O4(g) was placed in a 1.0 litre flask at 400 K, the following equilibrium is reached with a total pressure of 6 bar.N2O4⇌2NO2Assuming ideal behaviour of the gases, the partial pressure of N2O4(g) at equilibrium in bar is:[R= 0.083 bar L mol−1k−1]

A
3.32
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B
9.32
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C
0.66
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D
2.68
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Solution

## The correct option is B 0.66N2O4 ⟶ 2NO20.1−x 2x P1 = partial pressure of N2O4 P2 = partial pressure of NO2 Partial pressure of N2O4=6×(0.1−x)(0.1+x) Total moles = PVRT Total moles = 6×1(0.083×400) =0.18 moles Total moles =0.1+x=0.18x=0.18−0.1=0.08 Partial pressure of N2O4=6×(0.1−0.08)(0.1+0.08)Partial pressure of N2O4=(0.6−0.48)0.18Partial pressure of N2O4=0.66 barHence, option C is correct.

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