Predicting Spontaneity of 2CO(g)+O2(g)=2CO2(g)

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SUMMARY

The reaction 2CO(g) + O2(g) = 2CO2(g) has a ΔH of -566 kJ, indicating it is exothermic. The entropy change (ΔS) is negative due to a decrease in the number of moles from reactants to products, which results in a decrease in entropy. Therefore, this reaction is spontaneous only at low temperatures, as dictated by the relationship between ΔH and ΔS. The established trends confirm that for ΔH < 0 and ΔS < 0, spontaneity occurs at low temperatures.

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Homework Statement


Will the following processes be spontaneous only at high temperatures, only at only temperatures, at all temperatures, or at no temperatures?

2CO(g)+O2(g)=2CO2(g)
Delta(H)=-566kJ

Homework Equations

The Attempt at a Solution


So i just need to figure out how to predict the sign of Delta(S) but I am not sure how to do that with just the balanced equation given, any tips would be helpful.

Thanks!
 
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I think this question is referring to the effect of temperature on the equilibrium constant. For an exothermic reaction (negative delta H), does the equilibrium constant increase with temperature or decrease with temperature?

Chet
 
The total moles in the reactants are more than the products. This is decreasing entropy. If the total moles were more in the products, then this would be increasing entropy. So, when ΔH is negative, and ΔS<0 (decreasing entropy), we know that this is spontaneous at low temps.

The trends are as follows:
ΔS>0 and ΔH<0 Spontaneous (ΔG<0)
ΔS>0 and ΔH>0 spontaneous at high temps.
ΔS<0 and ΔH<0 spontaneous at low temps
ΔS<0 and ΔH>0 non-spontaneous (ΔG>0)

Notice that when the signs for both ΔS and ΔH are the same, it dictates the conditions of the reaction (H and S <0 are both spont. at low temps) and (H and S>0 are spont. at high temps). Hope this helps.
 

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