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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Oct 2007 19:21:08 IST
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Q. A mixture of SO2 and O2 at 1 atm in the mole ratio of 2:1 is passed thru a catalyst at 1170 (degree celsius) at a rate sufficient for the attainment of equilibrium.The existing gas,suddenly chilled and analysed , is found to contain 87% SO3 by volume. Calculate kp for the reaction: SO2+ O2 (reversible rxn.) SO3. SALUTES ASSURED TO ALL CORRECT ANSWERS.
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27 Oct 2007 20:08:38 IST
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Is the answer 2455.4
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28 Oct 2007 13:38:52 IST
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look,
let the initial concentrations of so2 & o2 be 2 & 1 litre....
& let us assume that x litre of o2 gets used up for attaining equi..
so final volumes will be:-
(2-2x) +(1-x) + (2x) = 3-x
now we r given 87% so we get
2x/3-x * 100= 87/100
calculate x from above....then its easy man.....calculate pressure & hence Kp.
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