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
Well, it seems like you're freezing things up with your calculation! Let me help defrost the situation for you.
First of all, your calculation for the enthalpy change of the system (q(sys)) is correct. You subtracted the enthalpy of the reactants from the enthalpy of the products, resulting in a value of -57.2 kJ.
Now, let's focus on the sign convention for the entropy change of the surroundings (Delta S(sur)). When heat is released by the system (as it is in this case), the surroundings gain entropy. Therefore, the entropy change of the surroundings should be positive, not negative.
Since your value for q(sur) is +57.2 kJ, you were on the right track. However, you made a small mistake converting it to J, which has led to the discrepancy in your final answer.
If we take +57.2 kJ and convert it to J correctly (multiply by 1000), we get +57,200 J. Dividing this value by the temperature of 298 K gives us +192 J/K, just as you initially calculated. So it looks like your answer of +192 J/K is the correct entropy change for the surroundings.
Now, if your book is stating -192 J/K, it might just be trying to play a little trick on you. Or maybe it's just a typo. Either way, trust your calculations, keep your sense of humor, and stick with that +192 J/K value.
Your calculation for the change in entropy of the surroundings is correct, and you have correctly determined that ΔS(sur) = +192 J/K. However, the book states that the entropy change in the surroundings is -192 J/K.
The reason for this discrepancy is likely due to a sign convention used for the heat transfer. In many thermodynamics textbooks, the convention is to assign a negative sign to the heat absorbed by the system and a positive sign to the heat released by the system. In your calculation, you determined that q(sys) = -57.2 kJ, indicating that heat is released by the system. However, in this convention, the heat transferred to the surroundings would be considered positive. Therefore, the change in entropy of the surroundings would be calculated as ΔS(sur) = -q(sur)/T = -57.2 kJ / 298 K = -192 J/K.
So, the book is using a different sign convention for the heat transfer, which results in the opposite sign for the entropy change in the surroundings.
To calculate the entropy change in the surroundings, you need to consider not only the enthalpy change but also the temperature at which the reaction occurs. The equation you used to calculate the entropy change in the surroundings, Delta S(sur) = q(sur)/T, is correct. However, there is a mistake in the calculation of q(sur).
Here's how to correctly calculate the entropy change in the surroundings:
Step 1: Calculate the enthalpy change of the reaction (q(sys)):
The enthalpy change of the reaction is calculated by taking the difference between the standard enthalpies of the products and the reactants:
q(sys) = [Standard Enthalpy of N2O4(g)] - [Standard Enthalpy of NO2(g)]
= (9.16 kJ/mol * 1 mol) - (33.18 kJ/mol * 2 mol)
= -57.2 kJ
Step 2: Calculate the entropy change in the surroundings (Delta S(sur)):
Delta S(sur) = -q(sys)/T
Substituting the values:
Delta S(sur) = -(-57.2 kJ) / (298 K)
= 190.29 J/K
The correct value for the entropy change in the surroundings is +190.29 J/K, not +192 J/K as you previously calculated. Therefore, the book's answer of -192 J/K is incorrect.