C6H12O6 (aq) = 602 (g) - 6CO2 (g) = 6H20 (l)

H -2837

A)Exo or endothermic?
B)write an expression for the equilibrium constant
C)Give the value of the equilibrium constant is very large, would this reaction be fast or slow?
D)effects of equilibrium if..
i. incerasing temp
ii. increasing pressure by decreasing volume
iii. decreasing concentration of oxygen
i.V. increasing the concentration of carbok dioxide
u. adding catalyst

A) To determine if the reaction is exothermic or endothermic, you need to consider the enthalpy change (∆H) of the reaction. The given value for ∆H is -2837 kJ/mol, which indicates that the reaction is exothermic. The negative sign implies that heat is released during the reaction.

B) To write an expression for the equilibrium constant (K), you need to know the balanced chemical equation and the stoichiometric coefficients of the reactants and products. The balanced equation is:
C6H12O6 (aq) + 602 (g) -> 6CO2 (g) + 6H20 (l)

The expression for the equilibrium constant would be: K = ([CO2]^6[H2O]^6) / ([C6H12O6][O2]^6), where the square brackets represent the concentrations of the species involved.

C) If the equilibrium constant (K) is very large, it indicates that the products are greatly favored over the reactants at equilibrium. This means that the reaction is proceeding almost completely in the forward direction. Therefore, this reaction would be fast.

D) Effects of changes on the system at equilibrium:
i. Increasing temperature: According to Le Chatelier's Principle, increasing the temperature will shift the equilibrium in the endothermic direction to absorb the extra heat. In this case, it would shift towards the reactant side, which means the concentration of C6H12O6 and O2 would increase, while the concentration of CO2 and H2O would decrease.

ii. Increasing pressure by decreasing volume: This reaction involves only gases, so changing the pressure affects the equilibrium. According to Le Chatelier's Principle, increasing the pressure would favor the side with fewer moles of gas. Therefore, in this reaction, it would favor the right side, resulting in an increase in the concentration of CO2 and H2O.

iii. Decreasing concentration of oxygen: Decreasing the concentration of O2 would shift the equilibrium towards the reactant side to compensate for the decrease. As a result, the concentration of C6H12O6 would increase, while the concentration of CO2 and H2O would decrease.

iv. Increasing the concentration of carbon dioxide: Increasing the concentration of CO2 would shift the equilibrium towards the product side to alleviate the extra concentration. Consequently, the concentration of CO2 would decrease, while the concentration of C6H12O6 and O2 would increase.

v. Adding a catalyst: Adding a catalyst would not affect the equilibrium constant or the concentrations of the reactants and products at equilibrium. Instead, it would increase the rate of both the forward and reverse reactions by providing an alternative pathway with lower activation energy. The equilibrium position remains the same, but the system reaches equilibrium faster.

Is this just to keep me busy? What do you not understand about this?

Remember -dH is exothermic.
Large Keq means the reaction is favorable; small K means not very favorable.
Le Chatelier's Principle says when a system at equilibrium is disturbed, it will shift so as to undo what we've done to it. Add reagent, it shift AWAY from the addition, reducing reagent it shift TO it, increase in P shifts to side with fewer gas mols.