September 11, 2024, 02:27:50 AM
Forum Rules: Read This Before Posting

### Topic: Calculating Standard Entropy of Vaporization at A Different Temperature  (Read 23946 times)

0 Members and 1 Guest are viewing this topic.

#### big

• Regular Member
• Posts: 79
• Mole Snacks: +1/-0
##### Calculating Standard Entropy of Vaporization at A Different Temperature
« on: March 19, 2012, 09:59:40 PM »
Calculate the standard entropy of vaporization of water at 85°C, given that its standard entropy of vaporization at 100.°C is 109.0 J/K*mol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/K*mol and 33.6 J/K*mol, respectively, in this range.

I realize that it is possible to do this problem by finding the entropy change to heat liquid water from 85°C to 100°C, finding the entropy change in vaporizing this liquid water (which would just be 109.0J/K*mol), and finding the entropy change to lower the temperature of the water vapor back to 85°C. The answer obtained from this method is 111 J/K*mol.

However, I initially did it a different way and wanted to know what was wrong with what I did:

I thought to find the :delta: Hvaporization at 100°C by doing 373.15K *109.0 J/K*mol = 40673.35 J/mol. Then I used Kirchoff's law to find :delta: Hvaporization at 85°C: :delta: HT2 = :delta: HT1 + (T2-T1)(:delta: CP) = 40673.35 J/mol + (358.15 K - 373.15 K)(33.6 J/K*mol - 75.3 J/K*mol) = 40673.35 J/mol + 625.5 J/mol = 41298.85 J/mol.

Finally, I took this :delta: Hvaporization at 85°C and divided it by 358.15 K to get (41298.85 J/mol)/(358.15 K)=115 J/K*mol.

As there's a 4 K difference between 115 K and 111 K, I didn't think it was due to rounding error. If someone could tell me what I did wrong, I would really appreciate it!

#### blaisem

• Regular Member
• Posts: 87
• Mole Snacks: +5/-0
##### Re: Calculating Standard Entropy of Vaporization at A Different Temperature
« Reply #1 on: March 25, 2012, 07:57:12 PM »
Ok, let's do this.  So, you know that  ΔS=Cpln(T2/T1)

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.

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) ΔQliquid/mol = CpliquidΔT

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) ΔQvapor/mol = CpvaporΔT

and (3) - (4) is

( 5) CpliquidΔT - CpvaporΔT = ΔT(Cpliquid - Cpvapor)

In step 2 from the initial method, we added ΔSvap at the normal temperature.  Here, we add in ΔHvap at the normal temperature, 109 J/K/mol * 373.15 K

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

ΔH85degC = ΔH100degC +  ΔT (Cpliquid - Cpvapor)

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 ΔHvap at the normal temperature, and subtract the enthalpy of cooling the vapor back down to 85 degC. At this point, you divide ΔH85degC 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 = CpΔT

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) = (Cp/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

ΔStotal = Integral (from T1 to T2) of Cp/T * ΔT

The integral solves to

Cp [ln(T2) - ln(T1)] = Cp*ln(T2/T1)

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.
« Last Edit: March 25, 2012, 08:13:53 PM by blaisem »