anonymous
  • anonymous
2SO2 (g) + O2 (g) 2SO3 (g) + heat. What will happen if more sulfur trioxide gas is added to the reaction mixture?
Chemistry
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
a-The equilibrium will shift toward the products. b-The equilibrium will shift toward the reactants. c-Keq will increase. d-Keq will decrease. e-There will be no shift in equilibrium.
anonymous
  • anonymous
@Missiey
Photon336
  • Photon336
Keq cant decrease or increase it's a constant value at a particular temp. you would have to change temp for this to happen. @bransonsmommy what do you think the answer is

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More answers

anonymous
  • anonymous
would it be E then? @Photon336
Photon336
  • Photon336
why do you think it's E
anonymous
  • anonymous
it was kind of a guess but if it cant increase or decrease it cant be C or D, but E just looked like the best guess to me, im not to good in chem :/
Photon336
  • Photon336
well @bransonsmommy it will have an effect. think about it like this the reaction is already at equilibrium, so there's a certain amount of reactants and products at equilibrium for this reaction. if we add more on one side, we call that a stress, the reaction has to adjust to that stress. follow so far?
anonymous
  • anonymous
somewhat, i'm leaning towards A, but i am still a little fuzzy on this...
Photon336
  • Photon336
@bransonsmommy \[Keq = \frac{ products }{ reactants } = \frac{ [C]^{}[D] }{ [A][B] }\]
Photon336
  • Photon336
first do you understand what equilibrium is? @bransonsmommy
anonymous
  • anonymous
not really.. im doing online classes and they give us no notes or anything..
Photon336
  • Photon336
ok
Photon336
  • Photon336
when you have a reaction, say you have a bridge right? the products are on one side the reactants are on the other. the reaction will proceed until it reaches a point where the rate of forward reaction = the rate of the reverse reaction, that's equilibrium a little analogy here is the number of cars going back and forth is the same. Equilibrium does not mean that the concentrations of the reactants and products are equal. Can you see why? Also what is favored in this reaction, reactants or products? at equilibrium |dw:1443623016742:dw|
anonymous
  • anonymous
i think i kinda get it a little better, so the answer would be B?
Photon336
  • Photon336
yeah i would say so
anonymous
  • anonymous
thank you!
Photon336
  • Photon336
but there's a lot to be learned here at least @bransonsmommy, sometimes they give you Keq value and sometimes they dont for this question they didn't. like if i gave you a keq of 10^6 at 25 C what type of reaction is this? how do you know what does it favor?

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