Question

Question 1

Glucose metabolism can be represented by the following chemical reaction:

C₆H₁₂O_6(aq)+6O_2(g)6CO_2(g)+6H₂O_(l)

Question 1 Glucose metabolism can be represented by the following chemical reaction: C6H12O6(aq)+6O2(g) deltaH for the reaction is -2837 kJ/mole. Is this reaction endothermic or exothermic? Write an expression for the equilibrium constant for this reaction. Given that the value of the equilibrium constant is very large, would you expect this reaction to be fast or slow? Explain the effect on equilibrium of Increasing temperature Increasing pressure by decreasing the volume Decreasing concentration of oxygen Increasing the concentration of carbon dioxide Adding a catalyst 6CO2(g)+6H2O(l)

$\rightleftharpoons \Delta$     H for the reaction is -2837 kJ/mole.

Is this reaction endothermic or exothermic?

Write an expression for the equilibrium constant for this reaction.

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

Explain the effect on equilibrium of

Increasing temperature

Increasing pressure by decreasing the volume

Decreasing concentration of oxygen

Increasing the concentration of carbon dioxide

Adding a catalyst

Answer 1

C6H1 206(aq) 6 02(g) 6 CO2 (g) 6 H2O(1) heat + <z) + + (a) since DH is negative -» heat is released => reaction is exothermic [Co2]A6 (b) Keq = [C6-1206] [02]AG (c) Reaction rate cannot be inferred from value of Keq (d) Increase temperat ure adding heat equilibrium shifts to left according to Le Chatelier 's principle Ce) No change in moles of gas -increasing pressure/decreasing volume has no effect on equilibriunm Cf) Decreasing concentration of o2 → equilíbriu shifts to left according to Le Chatelier's principle (g) Increasing concentration of co2 -equilibrium shifts to left according to Le Chatelier 's principle Ch) Adding catalyst has has no effect on equilibrium