Calculate the entropy change in the surroundings when 1.00 mol of N2O4(g) is formed from 2 mol NO2(g) under standard conditions?

Here's my approach. I get the right magnitude but the wrong sign.

1. Get the delta H of the reaction

2. Use that as the heat released to or from the surroundings. If Delta H is negative then heat is released to the surroundings?

3. Calculate Delta Ssur = qsur/T

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By combining

N2 + 2O2 > N2O4 Delta Hzero=9.16 kJ/mole

2NO2 > N2 + 2O2 Delta Hzero 2(-33.18)

Thus the net Delta H is -57.2 kJ for 1 mole of N2O4 formation.

Thus the reaction is exothermic and ADDS heat to the surroundings so I felt that

The book however says the answer is +192 ? Any suggestions


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