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For the reaction NOBr (g) hArrNO(g)+1/2Br_(2)(g),K_(p)=0.15 atm at 90°C. If NOBr, NO and Br_(2) are mixed at this temperature having partial pressures 0.5 atm,0.4 atm & 2.0 respectively, will Br_(2) be consumed or formed? |
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Answer» Solution :`Q_(P)=([P_(Br_(2))]^(1//2)[P_(NO)])/([P_(NOBR)])=([0.20^(1//2)[0.4]])/([0.50])=0.36` `K_(p)=0.15` HENCE, reaction will shift in backward direction, Therefor `Br_(2)` will be consumed. |
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