Without doing any calculations, determine the sign of Delta S(sys) and Delta S(surr) for the chemical...

Without doing any calculations, determine the sign of Delta S(sys) and Delta S(surr) for the chemical reaction below.

A) CO2(g) ---> C(s) + O2(g) Delta H(rxn) = +393.5 kJ

Explanation please.

Also, predict under what temperature (all temperautes, low tempertures, or high temperatures or none) the reaction will be spontaneous.

Expert Solution

from the reaction CO2(g)------>C(s)+ O2(g), the disorder remains the same since the no of gas molecules on reactant side are same as that on the product side. however, no of atoms in CO2 are more than that of O2. Hence more disorder in CO2 compared to O2. So there is a slight decrease in entropy during the course of reaction.

so entropy of system decreases

Since the entalpy of reaction is +ve, the reaction is endothermic. Hence the energy of surroundings increases and hence entropy of surroundings decreases.

since deltaG= deltaH-T*deltaS

deltaS of system is -ve, but not significant compared to deltaH. deltaH is always +ve. hence deltaG is always +ve at all temperatures. The reaction is non spontaneous at all temperautres.

queen_honey_blossom answered 2 years ago

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