Which statement is true?
a reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.
a reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
a reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.

A

Explanation:

The entropy of a system has to deal with how energy is stored in a system. It tends to looks at the degree of disorderliness in the system.

The spontaneity of a reaction deals with the possibilities of the reactions occurring, while an exothermic reaction is one in which energy is released to the surroundings, the change in enthalpy here is negative.

The change in entropy or the change in enthalpy cannot dictate if a reaction is feasible I.e spontaneous or not. What determines the spontaneity of a chemical reaction is the change in free energy. For a reaction to be spontaneous, this value must be negative.

The change in free energy, the enthalpy and entropy are connected by the equation below:

ΔG = ΔH - TΔS

Now back to the question, for the first option, we were told that entropy increased and system is exothermic meaning negative enthalpy and positive entropy. In this case , change in free energy will be negative over all temperatures and hence, the reaction is spontaneous.

To the second option, we have negative entropy and negative enthalpy. Here change in free energy will only be negative at low temperature, which of course is not the best answer here

To the third option, we have negative entropy and positive enthalpy. This means change in free energy is only negative at high temperatures which of course is not the best answer.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.