Question

Hi, I have some questions about the concepts of energy in chemistry.

#1 A particular chemical reaction has a negative deltaH and a negative deltaS. Which statement is correct?
The answer:
-The reaction becomes spontaneous as temperature decreases.
Can someone please explain to me how this is true?

#2 For the process O2(g)---> 2O(g), deltaH = +498kJ. What would be predicted for the sign of deltaS(rxn) and the conditions under which this reaction would be spontaneous?
The answer:
deltaS(rxn)=positive & spontaneous at low temperatures only.
Can someone please elaborate on this?

#3 For which reaction, carried out at standard conditions, would both the enthalpy and entropy changes drive the reaction in the same direction?
Answer:
N2(g) + 2O2(g) ---> 2NO2(g) deltaH=+67.7
Why?

Answer 1

All the three questions are related to the application of expression -

(i) Delta G = Delta H - T Delta S   

(ii) Making out the sign of Delta G ( note that a negative sign with Delta G represents spontaneous reaction )

#1. In this case a negative value of T ( due to decrease in temperature ) when multiplied with negative value of DeltaH would yield a positive value, but this is again multiplied with a negative sign ( as per above expression ) thus making the value negative.Again , since Delta H is negative , the result would be to give out negative value for DeltaG , thereby causing the reaction to be spontaneous with decrease in temperature.

#2 . In the given reaction ,

O2(g) ----------------> 2O (g) with a positive Delta H ( = 498kJ )

there happens to be an increase in entropy , because a molecule of oxygen is changed to two atoms of oxygen. So Delta S would be positive.

Since Delta H is also positive for this reaction , the reaction would be favoured at high temperature making Delta G more negative causing spontanity of the reaction . The given answer , therefore , calls for a check.

# 3. In the given reaction there is a decrease in entropy [ since three molecules of the gaseous reactants are converted to two molecules of a gaseous product , NO2 ] , making the product T Delta S less negative, hence

A decrease in Delta H ( ie. lower value of Delta H ) would yield a negative Delta G . causing reaction to occur in same direction.