Question

At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are present: [SO2] = 0.010 M; [SO3] = 10.M; [O2] = 0.010 M.Is the mixture at equilibrium? If NOT at equilibrium, in which direction (as the equation is written), left to right or right to left, will the reaction proceed to reach equilibrium?Yes, the mixture is at equilibrium.No, left to right.No, right to left.There is not enough information to be able to predict the direction.

Answer 1

Step-by-step explanation:

The given reaction is as follows.

`2SO_(2) + O_(2)(g) \rightarrow 2SO_(3)(g)`

Value of equilibrium constant is given as
`K_(c)` = 4.3 \times 10^{6}[/tex].

Concentration of given species is
`[SO_2]` = 0.010 M;
`[SO_3]` = 10.M;
`[O_2]` = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

Q =
`([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`

Putting the given concentration as follows.

Q =
`([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`

Q =
`((10)^(2))/((0.010)^(2)(0.010))`

Q =
`10^(8)`

It is known that when Q >
`K_(eq)`, then reaction moves in the backward direction.

When Q <
`K_(eq)`, then reaction moves in the forward direction.

When Q =
`K_(eq)`, then reaction is at equilibrium.

As, for the given reaction Q >
`K_(eq)` then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.