If a reaction is not spontaneous is there any of making that reaction occur?

[Ahmed Abdullah]

If a reaction is not spontaneous is there any way of making that reaction occur?

Question Details: We know that
del G=del H - T(del S)

if del H is positive and del S is negative then del G is always positive. There is no way of making del G negative. My question is whether such a reaction take place at all (to any extent)?
thx

 

[pkleinod]

if del H is positive and del S is negative then del G is always positive.

Not true! The changes in enthalpy and entropy refer to a single temperature and pressure. By changing the temperature or pressure, the free energy change will vary and may become negative. An example is the conversion of graphite to diamond. For this reaction (at room temperature and 1 atm.) the change in enthalpy is 450 cal. and the change in entropy is -0.79 cal/deg. But graphite and diamond have different densities, so that changing the pressure can effect a free energy change. Above a pressure of 22000 atm. diamond is more stable than graphite at room temperature.

 

[lightarrow]

Not true! The changes in enthalpy and entropy refer to a single temperature and pressure. By changing the temperature or pressure, the free energy change will vary and may become negative. An example is the conversion of graphite to diamond. For this reaction (at room temperature and 1 atm.) the change in enthalpy is 450 cal. and the change in entropy is -0.79 cal/deg. But graphite and diamond have different densities, so that changing the pressure can effect a free energy change. Above a pressure of 22000 atm. diamond is more stable than graphite at room temperature.

You could be right, but however seems strange to me that delta(H) of the reaction graphite --> diamond is positive; are you sure?

 

[lightarrow]

Question Details: We know that
del G=del H - T(del S)

if del H is positive and del S is negative then del G is always positive. There is no way of making del G negative. My question is whether such a reaction take place at all (to any extent)?
thx

If delta(G) of reaction is positive, the reaction can also happen, it only have a lower equilibrium constant:
when delta(G) < 0 then K >1, when delta(G) = 0 then k = 1, when delta(G) > 0 then k < 1:
delta(G) = -RTlnK
Example:
0.5N2 + O2 <--> NO2
delta(G) = 3.32*10^(4) + 1.57*10^(2)*T
at T = 1000K:
delta(G) = +3.48*10^(4) J/mol
but we all know how NO2 pollution by internal combustion engines (diesels) is a terrible world problem.

 

[pkleinod]

You could be right, but however seems strange to me that delta(H) of the reaction graphite --> diamond is positive; are you sure?

Hi Lightarrow. I'm not sure of exact value, which I took from "Thermodynamics" by Lewis and Randall, but I am pretty sure that graphite is more stable than diamond at standard temperature and pressure. A graph of the free energy change can be found at

www.physics.rutgers.edu/ugrad/351/Lecture 14.ppt

Why do you find it strange? Perhaps it is the power of advertising: "A diamond is forever!"

 

[Ahmed Abdullah]

If delta(G) of reaction is positive, the reaction can also happen, it only have a lower equilibrium constant:
when delta(G) < 0 then K >1, when delta(G) = 0 then k = 1, when delta(G) > 0 then k < 1:
delta(G) = -RTlnK

It finally eliminates my confusions.
Thx pkleinod.

 

[lightarrow]

Hi Lightarrow. I'm not sure of exact value, which I took from "Thermodynamics" by Lewis and Randall, but I am pretty sure that graphite is more stable than diamond at standard temperature and pressure. A graph of the free energy change can be found at

www.physics.rutgers.edu/ugrad/351/Lecture 14.ppt

Thanks for the interesting link.

Why do you find it strange? Perhaps it is the power of advertising: "A diamond is forever!"

I found it strange just for the fact that in diamond there are more covalent bonds per atom than in graphite (in diamond every C is bound to other 4 C atoms, in graphite to other 3 atoms).