Consider the following reaction at equilibrium: 2co2 (g) ↔ 2co (g) + o2 (g) ∆h° = -514 kj le châtelier's principle predicts that removing o2(g) to the reaction container will consider the following reaction at equilibrium: 2co2 (g) 2co (g) + o2 (g) ∆h° = -514 kj le châtelier's principle predicts that removing o2(g) to the reaction container will decrease the value of the equilibrium constant increase the partial pressure of co increase the value of the equilibrium constant decrease the partial pressure of co increase the partial pressure of co2

it will increase the amount of carbon monoxide hence will increase the partial pressure of CO.

Explanation:

The reaction is:

∆H° = -514

The reaction is

a) Exothermic

b) There are more gaseous moles on product side

c) carbon dioxide is the reactant and carbon monoxide and oxygen are products.

According to Le-Chatelier's Principle if we bring any change in the equilibrium reaction, the equilibrium shifts in the direction where it can nullify the change.

i) If we remove oxygen (product) then the reaction will shift in the forward direction.

It means it will increase the amount of carbon monoxide hence will increase the partial pressure of CO.

There will be no effect on the equilibrium constant.

The partial pressure of carbon dioxide will decrease.

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