InterviewSolution
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InterviewSolution
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Two gases in adjoining vessels were brought into contact by opening a stopcock between them. One vessel measured 0.25 litre and contained NO at 800 torr and 220 K, the other measured 0.1 litre and contained O_(2) at 600 torr and 220 K. The reaction to form N_(2)O_(4) (solid) exhausts the limiting reactant completely. (a) Neglecting the vapour pressure of N_(2)O_(4) what is the pressure of the gas remaining at 220 K after completion of the reaction ? (b) What weight of N_(2)O_(4) is formed ? (torr = mm) |
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Answer» Solution :Let us first calculate o. of moles of NO and `O_(2)` before the reaction tkes place. Let `n_(1)` and `n_(2)` be the no. of moles of NO and `O_(2)` respectively in each vessel. We have pV = nRT For NO : `(800)/(760) xx 0.25 = n_(1) xx RT , "" n_(1) = (200)/(760 RT)`. For `O_(2) : (600)/(760) xx 0.10 = = n_(2) xx RT, ""n_(2) = (60)/(760 RT)`. Since NO and `O_(2)` react in 2 : 1 molar RATIO `{:("(2NO",+,O_(2),rarr,N_(2)O_(4)")"),(" "(g),,(g),,(s)):}` `therefore (60)/(760 RT)` mole of `O_(2)` will react with `(120)/(760 RT)` mole of NO and so only `((200)/(760 RT) - (120)/(760RT))` mole, i.e, `(80)/(760 RT)` mole of NO shall remain after the completion of reaction. The pressure due to remaining NO can thus be calculated as : pV = nRT `p(0.25 + 0.1) = (80)/(760 RT) xx RT , p = 0.30` atm or 229 mm. Further, we know that `(120)/(760 RT)` mole of NO COMPLETELY changed to `N_(2)O_(4)`. `therefore` applying POAC for N atoms in `NO overset(O_(2))rarr N_(2)O_(4)` `1 xx` moles of NO = `2 xx` moles of `N_(2)O_(4)` `(120)/(760 RT) = 2 xx ("WT. of " N_(2)O_(4))/(92), "wt. of " N_(2)O_(4) = 402 g {{:("R = 0.082 LIT. atm K"^(-1) MOL^(-1)),(T = 220 K):}}`. |
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