Question

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

Answer 1

Answer;

increase the partial pressure of CO2

Explanation;

2CO(g) + O2(g) -------> 2CO2(g) + heat

• Removel of Oxygen makes the equilibrium to move towards the left side (reactants) because the reactants concentration is decreasing. According to the LeChâtelier’s principle equilibrium moves towards the lower concentration.
• When the partial pressure of any of the gaseous reactants or of the products is increased, the position of equilibrium is shifted so as to decrease its partial pressure. In this case, the removal of oxygen causes a shift to to the left which causes an increase in the partial pressure of CO2.

Answer 2

Answer:

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

Explanation:

The reaction is:

$2CO_{2}(g)--> 2CO(g) + O_{2} (g)$ ∆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.