For the equilibrium, equilibrium constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of these two gases be taken as represented by the point p in the figure, predict what will happen?

Question

For the equilibrium, equilibrium constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of these two gases be taken as represented by the point p in the figure, predict what will happen? (A) Immediately above equilibrium will be setup (B) Above reaction will go in the forward direction till it attains equilibrium (C) Above reaction will go in the backward direction till it attains equilibrium (D) Nothing can be said

Tardigrade Admin · Accepted Answer

From given information, Q, = tan 75Â° = 3.73=(conc. of iso-butane/conc. i of -butane) >Kc So, reaction will go in backward direction.

Q. For the equilibrium,
equilibrium constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of these two gases be taken as represented by the point $p$ in the figure, predict what will happen?

Immediately above equilibrium will be setup

Above reaction will go in the forward direction till it attains equilibrium

Above reaction will go in the backward direction till it attains equilibrium

Nothing can be said

From given information,
$Q, = t an 7 5^{°} = 3.73 = \frac{co n c . o f i so - b u t an e}{co n c . i o f - b u t an e} > K_{c}$
So, reaction will go in backward direction.

Q. For the equilibrium, equilibrium constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of these two gases be taken as represented by the point p in the figure, predict what will happen?