For the reaction NOBr (g)iffNO(g)+(1)/(2)Br_(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, and 0.2 atm respectively, will Br_(2) be consumed or formed?

For the reaction NOBr (g)iffNO(g)+(1)/(2)Br_(2)(g) K_(P)=0.15 atm at

1.	For the reaction NOBr (g)iffNO(g)+(1)/(2)Br_(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, and 0.2 atm respectively, will Br_(2) be consumed or formed?
Answer» <P> Solution :`Q_(P)=([P_(Br_(2))]^(1//2)[P_(NO)])/([P_(NOBR]])=([0.2]^(1//2)[0.4])/([0.50])=0.36` `K_(P)=0.15` `therefore Q_(P)gtK_(P)` Hence, reaction will shift in BACKWARD direction `therefore Br_(2)` will be consumed