chemistry

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I am a bit confused with this proble because the concentration of SO3 at equilibrium. Any help would be appreciated.

Given the reaction: SO2(g) + NO2(g) ⇌ NO(g) + SO3(g) ∆H = -42.6 kJ. How
will the concentration of SO3(g) at equilibrium be effected by the following:

a) Adding more NO2(g)
b) Removing some NO(g)
c) Increasing the temperature

  • chemistry -

    Students seem to get confused with Le Chatelier's questions but they are very simple to answer. Here is the secret. Le Chatelier's Principle says that the reaction will shift so as to undo what we do to it. In my own words, I reword that to say, just do the opposite.
    So, if we add NO2, it must shift to the right (so as to USE UP) the NO2 added. Shifting to the left won't use it up--it will create more.
    Removing NO will shift the reaction to the right (to produce more). Shifting to the left will use it up and that isn't what we want.
    I find it easier to rewrite the equation when T is involved to make it look more like the other two questions.
    SO2 + NO2 ==> NO + SO3 + heat
    Thus, adding heat just makes it go to the left so as to use up the heat we've added. Shifting to the right will produce heat and that isn't what we want.
    Here is a trick I've found most students get from the start. Underneath the arrow, draw a solid line and think of the solid line as a rope. You may pull the rope on either end--your choice. You want to pull the rope, on either end, to do the opposite of what's been done to the reaction.
    For adding NO2: We pull on the right end of the rope so as to drag more NO2 to us. That is the equivalent of shifting the reaction to the right. Removing NO: We pull on the right side of the rope so as to drag more NO our way which is the equivalent of shifting the reaction to the right. Adding heat: we pull on the left side of the rope to pull it our way. That means we are pulling the heat our way which reduces the heat and is the equivalent of shifting the reaction to the left.

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