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Assume that the decomposition of HNO(3) ...

Assume that the decomposition of `HNO_(3)` can be represented by the following equation
`4HNO_(3(g)) hArr 4NO_(2(g)) + 2H_(2) O_((g)) + O_(2(g))` and the reaction approaches equilibrium at 400K temperature and the copper turnning 0 atm pressure. At cquilibrium partial pressure of `HNO_(3)` is 2 atm. Calculate Kc in `("mole" //L)^(3)` at 400 K.

A

4

B

8

C

16

D

32

Text Solution

Verified by Experts

The correct Answer is:
D
`4HNO_(3(g)) harr 4NO_(2(g))+2H_(2)O_((g))+O_(2(g))`
`P_("Total")=P_(HNO_(3))+P_(NO_(2))+P_(H_(2)O)+P_(O_(2))`
`:. P_(NO_(2))=4_(PO_(2))` and `P_(H_(2)O)=2P_(O_(2))`
`:. P_("Total")=P_(HNO_(3))+7P_(O_(2))`
`implies 30-2=P_(O_(2)) xx 7, P_(O_(2))=4`
`Kp=(P_(NO_(2))^(4) xx P_(H_(2)O) xx P_(O_(2)))/(P_(HNO_(3))^(4))`
`=((4 xx 4) xx (2 xx 4)^(2) xx 4)/(2^(4))=2^(20)`
`Kp=Kc(RT)^(Delta n_((g)))=Kc(0.0821 xx 400)^(3)`
`implies Kc(0.08 xx 400)^(3) implies Kc=32`
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