I calculated the values I thought I needed, by using the information I found online:
Quote:
Free energy of O2 = 0 kJ/mol * 3 mol = 0 kJ
G of CH4 = -50.8 kJ/mol * 2 mol = -101.6 kJ
G of NH3 = -16.4 kJ/mol * 2 mol = -32.8 kJ
G of H2O = -237.14 kJ/mol * 6 mol = -1422.84 kJ
G of HCN = 124.7 kJ/mol * 2 mol = 249.4 kJ
G of products = -134.4 kJ
G of reactants = -1173.44 kJ
G = +1039.04 kJ (not spontaneous)
H of O2 = 0 kJ * 3 mol = 0 kJ
H of CH4 = -74.87 kJ * 2 mol = -149.74 kJ
H of NH3 = -45.90 KJ * 2 mol = -91.8 kJ
H of H2O = -285.8 kJ * 6 mol = -1714.8 kJ
H of HCN = 130.5 kJ * 2 mol = 261 kJ
Hr = 1212.26 kJ
S of O2 = 205.03 kJ * 3 = 615.09 kJ
S of CH4 = 186.25 * 2 = 372.5 kJ
S of NH3 = 192.34 * 2 = 384.68 kJ
S of H2O = 69.91 * 6 = 419.46 kJ
S of HCN = 201.67 * 2 = 403.34 kJ
Sr = 549.47 kJ
And then, I plugged the values into the Gibbs Equation:
Quote:
ΔG = ΔH-TΔS
1039.04=1212.26-T(549.47)
-173.22=-T(549.47)
0.31525=T
0.31525 K
But I'm honestly confused as to what the answer I obtained really means. Is this the temperature at which the reaction continues to be nonspontaneous
(above this temperature will it remain spontaneous?)?
[Edited on 5/20/2018 by Velzee]aga - 20-5-2018 at 12:17
Blown away !
They're all in the Gas phase unless the water is at 100C+
Got a nagging feeling that makes a difference to the thermodynamic calcs.
It'll be fascinating/an education to see your question answered.Geocachmaster - 20-5-2018 at 13:16
The biggest problem is the units of S˚f. Entropy is given as J/mol*K, while enthalpy and Gibbs free energy are in KJ/mol*K. Before using
∆G=∆H-T∆S, you must either convert entropy to KJ (what I usually do) or convert everything else to J.
The answer is actually 315K. If all of the thermodynamic information looked up was at standard conditions, I would expect to get 298 K.
The temperature that you are getting has nothing to do with changes in spontaneity. The answer of 315 K is saying that when ∆H and ∆S are 1212 KJ
and 549 J, the temperature must be 315 K in order for ∆G to be 1039 KJ.
If you want to find the transition temperature between spontaneous and non-spontaneous, you must use a different approach. 0 is the transition between
spontaneous (negative value) and non-spontaneous (positive value), so plug that in for ∆G. Solving for temperature again using the original ∆H and
∆S values gives the temperature needed for ∆G to be 0. At higher temperature than this ∆G will become negative and it will be favorable, and at
lower temperature ∆G will increase and not be favorable.
0=1212-T(549/1000)
T=~2,208
So the transition temperature is ~2,208 K
Note: I did not look up any of the values or check the ∆H or ∆S calculations, I assumed they were correct and instead focused on the root of the
problem.
[Edited on 5/20/2018 by Geocachmaster]aga - 20-5-2018 at 13:48
Note: I did not look up any of the values or check the ∆H or ∆S calculations, I assumed they were correct and instead focused on the root of the
problem.
Does that mean that the reactions/equilibrium only exist(s) at ~2000 C ?
I really like the last bit - so many people just Google instead of sticking their neck out.
Nice one.walruslover69 - 21-5-2018 at 11:52
The reaction is not thermodynamically spontaneous or favorable at any temp. When I did the calculations with my own values I found
H=~940Kj
S=~-164J
If enthalpy is positive and entropy is negative then it will not be spontaneous at any temp.aga - 21-5-2018 at 12:34
Good maths work.
For me, the fact that the reagents and products were all gasses, apart from the water, it just felt all wrong.Geocachmaster - 21-5-2018 at 13:37
Looking up the values in some books I have, I get a ∆S of -167 J (close enough) for the reaction with all gas phase reactants and products.
However, If the HCN and H2O are liquid, the ∆S becomes 725 J. ∆H values for gas phase and with liquid HCN/H2O were found to
be 940 and 1264 KJ, respectively. This suggests that the reaction would only be spontaneous at high temperature and high pressure (the pressure needed
to keep HCN and H2O liquid). I suspect that ammonia would liquefy too, but the change in ∆S from this would not be enough to make it
negative, so the reaction should still be spontaneous at high temperature and pressure.walruslover69 - 21-5-2018 at 18:49
le chatelier's principle would state that if the reactants are liquid and the products gas then increasing the pressure would shift the equilibrium to
the reactants. Meaning that the reaction would not be favored. The stand enthalpy and entropy values values also change at high temp and pressure so
doing approximations like this arnt always that accurate.
[Edited on 22-5-2018 by walruslover69]Geocachmaster - 22-5-2018 at 04:04
No, Le Chatelier’s principle states that a system at equilibrium will readjust to counteract a change done to the concentration of a chemical. If
one chemical was removed, the system would shift somewhat to make more of the chemical. However, If all of the products or reactants were removed from
one side, they can continue to react in the liquid phase. Liquid HCN and H2O can react to make gaseous products. Therefore one could
conclude that the products would be favored, because they are lost from the liquid phase. You can see that Le Chatelier’s principle doesn’t really
apply.
Quote:
...would shift the reaction to the products. Meaning the reaction would not be favored.
By the way, saying that the system shifts to the products is saying that the reaction is more favorable
∆H and ∆S values do change with temperature, but usually not significantly. ∆G is meant to be a useful approximation. Values close to 0 at high
temperatures are unreliable, but reasonably large or negative values can usually be trusted.walruslover69 - 22-5-2018 at 08:08
Ya that was a typo, i just corrected it. I meant to say that it favored the reactants.
At standard pressure the reaction is never favorable. The only way to potentially make it favorable is to lower the pressure, but lowering the
pressure would turn the HCN to gas and it would become entropicly unfavorable. Using standard values there is no combination of temperature and
pressure where the reactant is favorable.
temperature not as much but Varying the pressure can change the Enthalpy and entropy quite a lot and can really complicate things.
I stand by what is said about le chatelier's principle. increasing the pressure on a system with liquid reactants and gaseous products would shift the
equilibrium to the reactants.
