Equilibrium of water

[wsid]

2H2(g) + O2(g) -->2H2O(g)
Which of the following would decrease the proportion of water vapor at equilibrium?
a. increasing partial pressure of O2 gas
b. decreasing volume of container
c.reducing amount of catalyst
d. raising the temperature

The ExamKracker book says the answer is D...wouldn't raising the temperature make the forward reaction favorable? Why is the answer D?

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[SweetBurger]

The reaction is exothermic therefore if you increased the temperature you would favor the reverse reaction.

 

[NextStepTutor_1]

Remember for questions that refer to how a reaction equilibrium shifts, use le-chatelier's principle. In this case, as SweetBurger said, the reaction is exothermic, so heat is a product. So, for each of the answer choices see how the reaction would be affected.

A) Increasing O2 would drive the reaction forward since O2 is a reagent
B) Decreasing the volume drives the reaction forward since there are 3 moles of gas on the reagents and 2 moles of gas produced, so the products are favored in the low volume state
C) Catalysts have no effect on equilibrium constants
D) Increasing temp. increases a product of this reaction (heat), so the reaction shifts to the left, and we have less H2O

 

[premed1001]

what is the catalyst in that reaction?

 

[Teleologist]

The catalyst choice is just to throw you off. The amount of catalyst won't change the position of equilibrium but might change the kinetics of the reaction - i.e. how long it will take to reach equilibrium.

 

[premed1001]

if there is no catalyst and choices a and b will drive it forward, so by elimination it can only be d