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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

there is equilibrium in the flask

B

there reaction will proceed in the forward

C

the reaction will proceed in the backward direction

D

the partial pressure of NOBr finally will be 0.05 atm

Text Solution

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The correct Answer is:
A
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