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∆S(system) = ∆H(system)/T since (assuming constant pressure) ∆H(system)=q(system), and if heat were added to the system, then entropy would increase. (+q(system) --> +∆S(system)). ∆S(surroundings) = -∆H(system)/T since if heat from the surroundings were added to the system, then the entropy of the surroundings would decrease (+q(system) --> -∆S(surroundings)).
The equations are opposite for the system vs the surroundings because heat is being transferred from one to the other, so the entropies must be opposite values to reflect this.
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