Consider the reaction: 2NO(g)+Br2(g)⇌2NOBr(g) Kp=28.4 at 298 K In a reaction mixture at equilibrium, the partial...

Consider the reaction:
2NO(g)+Br2(g)⇌2NOBr(g) Kp=28.4 at 298 K
In a reaction mixture at equilibrium, the partial pressure of NO is 105 torr and that of Br2 is 132 torr .

What is the partial pressure of NOBr in this mixture?

Expert Solution

Given reaction is 2NO(g) + Br₂(g) ⇌ 2NOBr(g) :K_p = 28.4

The partial pressure of NO = 105 torr

The partial pressure of Br2 = 132 torr

2NO(g) + Br₂(g) ⇌ 2NOBr(g)

The equilibrium constant , K_p = p²_NOBr(g) / [ p²_NO(g) x p _Br2(g) ]

From this p²_NOBr(g) = K_p x [ p²_NO(g) x p _Br2(g) ]

Plug the above values we get p²_NOBr(g) = 28.4 x [ 105² x 132]

= 41.3 x 10⁶

p _NOBr(g) =

= 6429 torr

Therefore the partial pressure of NOBr in the mixture is 6429 torr

queen_honey_blossom answered 2 hours ago

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