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Question

One mole of N2O4(g) at 300 K and 1.0 atm is kept in a closed container. 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
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B
2.4 atm
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C
2.0 atm
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D
1.0 atm
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Solution

The correct option is B 2.4 atmLet's assume initial mass of N2O4 is 100 g. Therefore, after 20% decomposition, the mass of N2O4 left will be 80 g. N2O4(g)⇌2NO2(g)Initial10092 mol0=1.08 molAt equilibrium8092 mol2046 mol=0.86 mol=0.43 mol From ideal gas equation, At 300 K,P0V=n0RT0 1×V=1.08×R×300 ...(1) At 600 K,P1V=n1RT1 P1×V=(0.86+0.43)×R×600 ...(2) Dividing equation (2) by (1), P11=1.29×6001.08×300 P1=1.29×21.08=2.38 atm≃2.4 atm

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