Question

Calculate the entropy change in the surroundings when 1.00 mol N2O4(g) is formed from 2.00 mol NO2(g) under standard conditions at 298 K.

I get +192 J/K. The book says -192 J/K. Here is my work, where am I going wrong?

Standard Enthalpy of NO2(g) = 33.18 kJ/mol * 2 mol = 66.36 kJ
Standard Enthalpy of N204(g) = 9.16 kJ/mol * 1 mol = 9.16 kJ
q(sys) = Enthalpy change = 9.16 kJ - 66.36 kJ = -57.2 kJ
q(sur) = -q(sys) = +57.2 kJ
Delta S(sur) = q(sur)/T = +57.2 kJ / 298 K = +192 J/K

Answer 1

you got qsystem and qsurr confused. qsurr is your change in enthalpy: -57.2KJ and qsy is +57.2 KJ