Q.

37.8 g N2O5 was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K
2N2O5(g)2N2O4(g)+O2(g)
The total pressure at equilibrium was found to be 18.65 bar.
Then, Kp = _______ × 10–2 [nearest integer]
Assume N2O5 to behave ideally under these conditions
Given : R = 0.082 bar L mol–1 K–1

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answer is 962.

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Detailed Solution

 Initial pressure of N2O5=37.8108×0.082×5001=14.35bar

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PTotal  at eqb =14.35+P=18.65P=4.3PN2O5=5.75barPN2O4=8.6barPO2=4.3barkp=(8.6)2×(4.3)(5.75)2=9.619=x×102x=961.9962

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