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`

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» `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

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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?

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