The following reaction is reactant-favored at equilibrium at room temperature.
COCl2(g) = CO(g) + Cl2(g)
Will raising or lowering the temperature make it product-favored?

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  1. We need to know if the reaction is endothermic or exothermic. We can calculate dH for the reaction through the heats of formation as follows:
    dHorxn = (n*dHo products) - (n*dHo reactants)
    Look in your text/notes and find the dHo vaues for CO, COCl2 and Cl2. I've done that below.
    n = 1; COCl2 = -223 kJ/mol
    n = 1; CO = -110.5 kJ/mol
    n = 1; Cl2 = 0
    Substitute into the equation above as follows:
    dHorxn = (-110.5 + 0) - (-223)
    dHorxn = -110.5+223 = + ?
    So we know the reaction is endothermic and we can rewrite the equation as
    COCl2 + heat ==> CO + Cl2.

    If we raise T it shifts the equilibrium to the right. Whether it is shifted enough to make it product favored depends upon delta G. To calculate dG,
    Do the same thing I've done above and calculate dSo rxn. With that information you can calculate dG.
    dG = dH - TdS.
    Set dG = 0, plug in dH and dS and calculate T at which the reaction is spontaneous to the right. I hope this makes sense to you. You may just be looking for raising T will make it product favored. My rationale is that raising T will shift the reaction to the right; however, whether that makes it FAVORED is true in my mind only if it makes the reaction spontaneous.

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