A reaction has (delta) H rxn = +107kJ and (delta) S rxn = 285 J/K. At what temperature...

At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?

Disclaimer: This is NOT a homework problem, but practice problem 17.2 from Tro, "Chemistry: A Molecular Approach." The given answer is: 375 K. I'm not sure I set the problem up correctly. Thanks for your help.

pg 783 Tro 2nd ed shows it as -107kJ.. .not positive.

anyway...
if ΔSuniverse >=0, the reaction is spontaneous.. pg 776
ΔSuniverse = ΔSrxn + ΔSsurroundings.. pg 780

so that...
0 <= ΔSrxn + ΔSsurroundings
-ΔSsurroundings <= ΔSrxn
ΔSsurroundings >= -ΔSrxn
ΔSsurroundings >= -285J/K

and we know that
TΔS = ΔH.. pg 783..

so that..
T = ΔH / ΔS
T = - 107000 J / (-285J/K) = 375K

*******
make sense or no?

welcome

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Try:
DS(surroundings) = DS(system)

DH/T = DS

Be sure you convert all units to J or to kJ.

107 kJ = 1.07X10^5 J. So,

1.07X10^5 / T = 287 J/K

T = 1.07X10^5 / 287 = 375 K