20. The following thermodynamic data at 25°C have been tabu- lated for the gas-p

Question

20. The following thermodynamic data at 25°C have been tabu- lated for the gas-phase reaction shown as follows (dotted lines represent hydrogen bonds): O...H-O 2 H-C C-H O-H HCOOH(g) (HCOOH)2(8) a. Write the equation for the standard heat of formation at (kJ mol 1) 362.63 785.34 (JK1 mol 1) 251.0 347.7 25°C of HCOOH(g) (specify the pressure and phase for each component) b. Calculate .Ho. AS®, and ,Go for the gas-phase dimer- ization at 298 K. Is the formation of dimer from mono- mers spontaneous under these conditions? Calculate the enthalpy change per hydrogen bond formed in the gas phase. Why is not a similar calculation useful to estimate the entropy or free energy of hydrogen-bond formation? c.

Explanation / Answer

The values on the left side correspond to gibbs and the values on the right side corresponds to entropy

you have too apply the equation

G = H – T*S, H is enthalpy, T is temperature , S is entropy

For HCOOH

G = H – T*S

H = G + T*S

H = -362.63 + (298.15 * 0.251) = -287.8 KJ/mole, remember to change entropy from J / mole to KJ / mole, do this dividing by 1000, this is the heat of formation for HCOOH, this is at 25C and 1 atm

repeat this for the other one (HCOOH )2

H = G + T*S

H = -785.34 + (298.15 * 0.3477)= -681.67 KJ / mole

For the reaction

2HCOOH ===== H2C2O4H2

To calculate enthalpy of reaction

G rxn = G products - G reactants

G products = n products * Gformation

G reactants = n reactants * Gformation

G products = -785.34 KJ / mole

G reactants = 2 * -362.63 = -725.26 KJ / mol

G rxn = products - reactants = -785.34 - (-725.26) = -60.08 Kj / mole

For enthalpy

H rxn = H products - H reactants

H products = n products * Hformation

H reactants = n reactants * Hformation

H rxn= -681.67 KJ / mole - (2 *  -287.8 KJ/mole) = -106.07 KJ / mole

The same goes for entropy

S rxn = S products - S reactants

Sproducts = 0.3477

S reactants = 0.251

S rxn = 0.3477 - 2*0.251 = -0.1543 KJ / mole or - 154.3 J / K mole

C) we said that the enthalpy for this reaction is -106.07 KJ / mole

I see 2 hydrogen bonds (dotted lines) are formed so enthalpy per bond is

-106.7 / 2 = -53.05 KJ/mole

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