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Predict whether the equilibrium 6CO2 (g) + 6H2O (l) <---> C6H12O6 (s) + 6O2 (g)...

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spock5 | Student, Undergraduate | (Level 1) Honors

Posted May 15, 2013 at 4:06 PM via web

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Predict whether the equilibrium

6CO2 (g) + 6H2O (l) <---> C6H12O6 (s) + 6O2 (g)

ΔH° = 2801.69 kJ/mol

would shift to the right, shift to the left, or remain unchanged if a) CO2 were decreased b) Pressure of O2 were increased c) One half of the glucose were removed d) The total pressure were decreased e) The temperature were increased

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mvcdc | Student, Graduate | (Level 1) Associate Educator

Posted May 15, 2013 at 4:30 PM (Answer #1)

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This problem is an application of Le Chatelier's Principle.

Le Chatelier's Principle tells us that if a system in chemical equilibrium experiences a change in concentration, temperature, volume, or partial pressure, the equilibrium will shift to reduce the stress.

a) Concentration of carbon dioxide is decreased:

Carbon dioxide is a reactant. Hence, we are decreasing the amount of reactants. The effect of this would be to reverse the reaction. The reverse reaction will produce more of the (decreased) product to reduce the stress. Equilibrium will shift to the left.

b) Pressure of O2 is increased:

Since the only the pressure of oxygen is increased, its partial pressure will increase. This is equivalent to increasing the concentration of oxygen. Hence, using the same logic used in (a), we need to produce more of the products (since oxygen is a product). Equilibrium will shift to the right.


c) Half of the glucose is removed:

This will decrease the concentration of glucose. However, since glucose is a solid, it does not affect equilibrium position. Pure liquids and pure solids do not change the equilibrium position. Equilibrium position will remain unchanged.


d) Total Pressure was decreased:

Decreasing the total pressure does not affect equilibirum. Equilibrium is affected by volume and the partial pressure only. A decrease in total pressure would also correspond to corresponding decreases in the partial pressure of the constituent gases. Equilibrium position will remain unchanged.


e) Temperature was increased:

Note that the reaction is endothermic, since enthalpy change associated with it is positive. This means that it is favorable for the forward reaction (the endothermic reaction). Another way to visualize this is to consider heat as a reactant (since it is endothermic). Thus, increasing temperature is increasing the reactants so it will favor formation of products (forward). Equilibrium will shift to the right.


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