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Mass action ratio or reaction quotient Q...

Mass action ratio or reaction quotient Q for a reaction can be calculated using the law of mass action `A(g)+B(g) harr C(g) + D(g) Q=([C][D])/([A][B])`. The value of Q decides whether the reaction is at equilibrium or not. At equilibrium, Q = K. For non equilibrium process, `Q ne K`. When `Q gt K`, reaction will favour backward direction and when `Q lt K`, it will favour forward direction.
In a reaction mixture containing `H_2`, `N_2` and `NH_3` at partial pressure of 2 atm, and 3 atm respectively, the value of `K_p` at `725 K` is `4.28 xx 10^(-5) atm^(-2)`. In which direction the net reaction will go ?
`N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g))`

A

Forward

B

Backward

C

No net reaction

D

Direction cannot be predicted

Text Solution

Verified by Experts

The correct Answer is:
B
`Q_(P)=(P_(NH_(3))^(2))/(P_(H_(2))^(3) xx P_(N_(2)))=((3)^(2))/((2)^(3) xx 1)=(9)/(8) gt K_(P)`
`Q gt K_(P) implies` backward direction
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