A question on thermodynamic heat transfer
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A question on thermodynamic heat transfer

[From: ] [author: ] [Date: 11-06-10] [Hit: ]
delta G = -(8.3145 J/molK)(303.15 K)ln 10 = - 5803.8 J/molNow, delta H changes to a very small extent at different temps but in order to solve for an approximate value which is close enough for all intensive purposes, we can solve two simultaineous equations utilizing the Gibbs-Helmholz thermodynamic equation which is as follows: delta G = delta H - T(delta S)So,......

delta G = - R T ln K

where delta G is the change in free energy of hydrolysis that occurs,
R is the gas constant which is expressed as: 8.3145 Joules/mol K
T is the absolute temperature where in this problem its (25 + 273.15) K = 298.15 K
ln K is the natural log of or log base e (not base 10) of the equilibrium constant

so delta G = - (8.3145 J/molK)(298.15 K)(ln 3066.48) = - 19891.9 J/mol = - 19.9 kJ/mol

Please note that the free energy value is negative which implies thermodynamically that this actual reaction is favorable to proceed forward or in other words considered either spontaineous overall or a downhill energy reaction.


For question 2, you first need to calculate the delta G value from the two different temperatures. So:

at T = 293.15 K, delta G = - (8.3145 J/molK)(293.15 K)ln .5 = 1689.5 J/mol
at T = 303.15 K, delta G = -(8.3145 J/molK)(303.15 K)ln 10 = - 5803.8 J/mol

Now, delta H changes to a very small extent at different temps but in order to solve for an approximate value which is close enough for all intensive purposes, we can solve two simultaineous equations utilizing the Gibbs-Helmholz thermodynamic equation which is as follows:

delta G = delta H - T(delta S)

So, at T = 293.15 K, 1,689.5 J/mol = delta H - 293.15 K (delta S)
and at T = 303.15 K, -5,803.8 J/mol = delta H - 303.15 K (delta S)

After solving the above two equatins simultaineously, we get the following values for delta H and delta S.

That is: delta H = 221,355.6 J/mol = 221.4 KJ/mol and delta S = 749.3 J/mol

Note that the reaction no matter what the temperature is greater than zero or endothermic since delta H is greater than zero or positive. The change in entropy or delta S is positive implying that this reaction goes from a more ordered state to a more disordered state. Also note that the change in free energy does change greatly between the two different temperatures implying also that at the lower temp, this reaction is not spontaineous and requires a much more energy to make it go forward than the reaction taking place at the higher temp which is considered thermodynamically favorable or spontaineous.
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