Chemistry

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Given the equilibrium: CO + 2H2 = CH3OH
H=-18.0kJ
How will the concentration of CO at equilibrium be affected by the following?

a) Adding more CH3OH
b) Removing some H2
c) Reducing the temperature.

I don't understand how i'm supposed to figure these out.. please help!

  • Chemistry -

    The long explanation is that when a system at equilibrium is put under stress (T, P, C, catalyst, etc), the system will shift in such as way so as to relieve the stress. Here is the simple explanation.
    A system at equilibrium will try to undo what we do to it.
    So let me rewrite the equation and I will ASSUME that all of the reactants and products are gases.
    CO + 2H2 = CH3OH + heat.
    Note that I have added the heat term into the equation instead of two stand alone items. Another note. There are only three answers to problems like this.
    a. The reaction shifts to the left.
    b. The reaction shifts to the right.'
    c. No effect (the reaction doesn't shift either way). The only time you see c is when the number of moles of gas are the same on both sides of the equation.
    (a) Adding more CH3OH.
    So adding CH3OH, the system will try to undo that. What must it do to undo what I did by adding CH3OH? It will try to take away CH3OH. How can it do that? The equilibrium can shift to the LEFT because that uses up CH3OH and forms more CO and more H2. (Of the three answers, you KNOW it can't shift to the right because that forms MORE CH3OH. You KNOW it can't stay the same because the moles are not the same on both side. The only answer that makes sense is a), shift to the left because that uses up the added CH3OH.
    b)Removing some H2. So we take away some H2. The system will try to undo that. So it will try to ADD H2. The ONLY way it can do that is to shift to the left. Shifting to the right would use up H2.
    I'll leave c for you. Reducing T means taking away heat or making it colder. So the reaction will ......

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