Glucose Metabolism can be represented by the following chemical reaction: C6H12o^(aq) + 6 02 (g) -------------- 6C02...

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Chemistry

Glucose Metabolism can be represented by the following chemicalreaction:
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 verylarge, 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

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3.9 Ratings (697 Votes)

The enthalpy change for the given reaction is 2837kJmol Since the enthalpy change is negative ie energy is liberated during the reaction Hence the reaction in exothermic A B The reaction is C6H12O6aq 2O2 6 CO2 6H2O l the equilibrium constant K is given by K CO26H2O6 C6H12O6O22 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 See Answer

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