[fl0raliessc]

Hello,

I was wondering if someone can help me figure out if I am understanding this problem correctly. My work is shown below the problem.

Given the following data, calculate the normal boiling point for formic acid (HCOOH).
                 DH°f (kJ/mol)   S° (J/mol K)
   HCOOH(l)       -410.       130.
   HCOOH(g)       -363       251

A. 2.57 K
B. 1730°C
C. 388°C
D. 82°C
E. 115°C

Here’s my work…
The reaction would probably be:
HCOOH (l)---> HCOOG (g)

I assume at normal boiling point, the liquid phase and solid phase will be at equilibrium.
Here we need to use the relation DG^0=DH^0-TDS^0

At equilibrium, the opposing driving forces are just balanced that is DG^0=0 and the liquid and gaseous phases coexist. That is the normal boiling point.

I know that in the example for determining the normal boiling point of Br2…when the givens are:
Br2(l)----> Br2(g), DH^0=31kj/mol and DS^0=93.0 J/K

DG^0=DH^0-TDS^0
0=DH^0-TDS^0
DH^0=TDS^0

T=(DH^0/DS^0)= (3x10^4 J/mol)/(93.0 J/K.mol) = 333 K

The normal boiling point for Br2= 333K

But for my problem, I need to get the values for H^0 and S^0 for both the phases, which are given serarately. How do I calculate DH^0 and DS^0, do I just subtract? Is there another way to look at this problem?

Thanks for your time!

 

[sdekivit]

we have:

HCOOH(l) --> HCOOH(g)

delta H = delta H(0)(g) - delta H(0)(l)

--> delta H = -363 + 410 = 47 kJ

delta S = delta S(0)(g) - delta S(0)(l)

--> delta S = 251 - 130 = 121 J/(mol K)

delta G = delta H - TdeltaS

--> T = 47 x 10^3/121 = 388 K

[fl0raliessc]

So I was correct. Thank you so much for the quick response! I really appreciate it!