Question

Glucose Metabolism can be represented by the following chemical reaction:

C6H12o^(aq) + 6 02 (g) -------------- 6C02 (g) + 6 H20 (i)

H for the reaction is -2837 kj/mole.

A: Is this reaction endothermic or exothermic?

B: Write an expressin for the equilibruim constant is very large, would you expect this reaction to be fast or slow?

C: Given that the value of equilibrium constant is very large, would you expect this reaction to be fast or slow?

D: Explain the effect on equilibrium of

: Increasing temperature

: Increasing pressure by decreasing the volume

: Decreasing concentration of oxygen

: Increasing the concentration of carbon dioxide

Answer 1

The enthalpy change for the given reaction is -2837kJ/mol.

Since the enthalpy change is negative i.e. energy is liberated during the reaction. Hence, the reaction in exothermic (A)

(B) The reaction is C₆H₁₂O₆(aq) + 2O₂ <-------------> 6 CO₂ +6H₂O (l)

the equilibrium constant K is given by

K= [CO2]⁶[H2O]⁶/ [C6H12O6][O2]²

as glucose and water are solid liquid and does not affect the reactant at equilibrium, it is disregarded and the values is kept at 1.

Therefore K= [CO2]⁶/ [O2]² --- 1.1

(C) If the value of equilibrium constant is large i.e. greater than 1, then the reaction is expected to have more products hence is fast. If we look at the equilibrium expression (1.1), the value of K will be greater than 1 if the concentration of CO2 and H2O is more than O2. Hence, the reaction will consist more of the products and less of the reactant.

(D) For this section, Le Chateliers principle will be used to explain the changes occuring. It states that if the dynamic equilibrium is disturbed by changing conditions then the position of the equilibrium shift in a manner to counteract the change.

(i) If one increases the temperature, then the position of the equilibrium changes. K is normally constant at constant temperature. As per the Le Chateliers principle, if we change a condition then the equilibrium will shift in a way to undo the change. Since the given reaction is an exothermic reaction, the change in temperature will shift the equilibrium toward the back reaction i.e towards the reactants as the forward reaction is exothermic.

(ii) If the pressure is increased, then again as per Le Chatelier's principle the position of the equilibrium will shift to reduce the total pressure by decreasing the total molecules produced. But, in this case, the total molecules on either side for the gaseous reactants is same, there will be no change in the position of the equilibrium.

(iii) If the concentration of oxygen is decreased, then the position of the equilibrium will shift towards left. That is more of products will react to form O2.

(iv) If the concentration of CO2 is increased, then the equilibrium will again shift towards left to produce more O2