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?

Explanation

`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.