The standard entropy change for 1.00 moles of Ar as it evaporates at -185.7 °C (∆H°-185.7 = 1558 cal mol-1) is?

1. 17.8 J mol-1K-1 2. 87.5 J mol-1K-1 3. 15.0 J mol-1K-1
4. 46.0 J mol-1K-1 5. 74.5 J mol-1K-1 6. none of the previous answers

I thought I would use deltaG=deltaH-TdeltaS, but I don't have deltaG, what do I do, or is deltaG zero because it is standard? Thank you. Or do I just used a table of entropy values to calculate this because it is standard?

I think I can answer your question but your phrasing is bad. You don't ask a question, really, and that word "is" is out of place.

Delta G is zero but not because it is standard but because it is in equilibrium at the boiling point (I assume that is the b.p. for Ar). Delta G is always zero when the system is at equilibrium such as the b.p. or m.p.

Sorry, but I didn't write the question, my teacher did.

I did deltaH/T=deltaS, and I got 17.81cal/Kmol
Is the conversion factor for cal to J 4.1858? If so I used that and got 74.5 (answer 5?)

To determine the standard entropy change (∆S°) for the evaporation of argon (Ar) at -185.7 °C, you can calculate it using standard enthalpy change (∆H°) and temperature (∆T):

1. Start by converting the given temperature from Celsius to Kelvin. To do this, add 273.15 to the value.
- In this case, -185.7 °C + 273.15 = 87.45 K

2. Using the equation ∆G° = ∆H° - T∆S°, rearrange it to solve for ∆S°:
- ∆S° = (∆H° - ∆G°) / T

3. Since you don't have the standard free energy change (∆G°), you can assume it is zero because the problem states "standard entropy change." In standard conditions, the free energy change is assumed to be zero.
- Therefore, ∆G° = 0

4. Substituting the values into the equation, you get:
- ∆S° = (∆H° - 0) / T = ∆H° / T

5. Now, plug in the values you have:
- ∆S° = (1558 cal mol-1) / (87.45 K) = 17.8 cal mol-1 K-1

6. Convert the result to the desired units (J mol-1 K-1) by using the conversion factor: 1 cal = 4.184 J.
- ∆S° = 17.8 cal mol-1 K-1 × 4.184 J cal-1 = 74.43 J mol-1 K-1

7. Rounding to the appropriate number of significant figures, the answer is approximately 74.4 J mol-1 K-1.

Therefore, the correct answer to the question is option 5: 74.5 J mol-1 K-1.