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The partial pressures of NO, Br2, and NO...

The partial pressures of `NO`, `Br_2`, and `NOBr` in a flask at `25^@C` are `0.01, 0.1`, and `0.04atm`, respectively. If the equilibrium constant at `25^@C` for the reaction
`2NO(g)+Br_2(g)hArr 2NOBr(g)`
is equal to `160atm^(-1)`, then we can say that

A

the partial pressure of `NOBr` finally will be `0.05 atm`

B

there is equilibrium in the flask

C

the reaction will proceed in the forward direction

D

the reaction will proceed in the backward direction

Text Solution

Verified by Experts

The correct Answer is:
B
We calculate the value of reaction quotient `(Q_P)` and compare it with the known value of `K_p` to predict the direction of the reaction that leads to equilibrium.
`Q_P=(P_(NOBr)^2)/(P_(NO)^2P_(Br_2))=((4xx10^(-2))^2)/((10^(-2))^2(10^(-1))`
`=160`
Since `Q_P` is equal to `K_P`, the system is at equilibrium.
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