Ok, let's do this. So, you know that ΔS=C_{p}ln(T_{2}/T_{1})

We have to:

1. sum the change in entropy from 85 degC to 100 deg C

2. add the change in entropy at 100 degC

3. subtract the change in entropy as the vapor is cooled from 100 deg C to 85 deg C.

This is your first method, and it is the correct answer.

Your second method is the exact same process, except you deal in enthalpy. To prove this, you may remember from Gen. Chem 1,

(1) ΔQ=mCΔT

In this case, we are talking about the molar enthalpy, so the mass in (1) is simply 1 mole, and it is divided through to the left side, and we end up with

(2) ΔQ/mol=CΔT

Our C is the heat capacity at constant pressure.

So, following method 1, but using enthalpy, we take at 358.15 K (85 degC)

(3) ΔQ_{liquid}/mol = C_{pliquid}ΔT

However, we are also going to subtract the change in enthalpy of the vapor going from 100 degC to 85 degC (step 3 in method 1). Thus:

(4) ΔQ_{vapor}/mol = C_{pvapor}ΔT

and (3) - (4) is

( 5) C_{pliquid}ΔT - C_{pvapor}ΔT = ΔT(C_{pliquid} - C_{pvapor})

In step 2 from the initial method, we added ΔS_{vap} at the normal temperature. Here, we add in ΔH_{vap} at the normal temperature, 109 J/K/mol * 373.15 K

Combining this and equation (5), we have Kirchhoff's Law,

ΔH_{85degC} = ΔH_{100degC} + ΔT (C_{pliquid} - C_{pvapor})

So, you can see that your method is the exact same process as method 1, where you add the change in enthalpy from 85 degC up to 100 degC, add the ΔH_{vap} at the normal temperature, and subtract the enthalpy of cooling the vapor back down to 85 degC. At this point, you divide ΔH_{85degC} by the temperature (358.15 K) to get the entropy. Why doesn't this work?

It doesn't work because, if you noticed in the process we took, we calculated the change in enthalpy over a range of temperatures. We cannot divide by simply one temperature to get the answer we want. Instead, we have to divide the enthalpy by the temperature at each step of this process to get ΔS at each degree Kelvin, and sum these ΔS to get the total ΔS that we are looking for.

From before, we know that

ΔH/mol = C_{p}ΔT

Now, we divide by T at this point, where T is whichever temperature we are at in the process of building up to 100 deg C and going back down to 85 degC. The expression for this is

ΔS = (ΔH/T) = (C_{p}/T)ΔT

We are summing each ΔS at each degree temperature to find the total ΔS, which leads us to an integral bounded by the range of our minimum temp and maximum temp. So

ΔS_{total} = Integral (from T1 to T2) of C_{p}/T * ΔT

The integral solves to

C_{p} [ln(T_{2}) - ln(T_{1})] = C_{p}*ln(T_{2}/T_{1})

Which is the equation you used in your first method. So, the reason why your answer was off is because you cannot assume one temperature when you divide by T, you have to divide by the total range of temperatures, which involves you summing up all of the ΔH/T values, which is the same thing as method 1. In other words, method 2 is incomplete. Method 1 solves the problem.