Question

Consider the following system at equilibrium at 1150 K: 2SO3(g) 2SO2(g) + O2(g) When some SO2(g) is removed from the equilibrium system at constant temperature: The reaction must: A. Run in the forward direction to restablish equilibrium. B. Run in the reverse direction to restablish equilibrium. C. Remain the same. Already at equilibrium. The concentration of O2 will: A. Increase. B. Decrease. C. Remain the same.

Answer 1

Answer:

The reaction must run in the forward direction to reestablish equilibrium.

The concentration of $O_{2}$ will increase.

Explanation:

Chemical reaction: $2SO_{3}(g)\rightleftharpoons 2SO_{2}(g)+O_{2}(g)$

Equilibrium constant in terms of concentration is expressed as:

                                       $K_{c}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$

$K_{c}$ is constant for a particular temperature.

When $SO_{2}$ is removed from system then $K_{c}$ decreases. Hence more $SO_{3}$ will decompose to produce more $SO_{2}$ and $O_{2}$ and keep $K_{c}$ constant in accordance with Le-chatelier principle.

Therefore the reaction must run in the forward direction to reestablish equilibrium.

Also, the concentration of $O_{2}$ will increase.