Is there a relationship between \Delta H_{sys} and \Delta S_{surroundings}?

[erty]

According to the 2nd law of thermodynamics, a spontaneous process will occur (or rather: is very likely to occur) if \Delta S > 0.

A chemical process occurs if \Delta G < 0, where G = H - TS.

Example:
H = -100 kJ
T = 1 K
S = -10 kJ/K
so \Delta G = - 190 kJ. In this example, \Delta G < 0 but \Delta S < 0.
Doesn't this contradict the 2nd law?

 

[dextercioby]

I don't see an S_{1} and an S_{2} so how did you compute \Delta S ?

Daniel.

 

[erty]

I don't see an S_{1} and an S_{2} so how did you compute \Delta S ?

Daniel.

Okay, then
\Delta H = -100 \mbox{kJ}
T = 1 \mbox{K} and
\Delta S = -10 \mbox{kJ/K}
because \Delta G = \Delta H - T\Delta S where T is constant.

 

[vanesch]

Okay, then
\Delta H = -100 \mbox{kJ}
T = 1 \mbox{K} and
\Delta S = -10 \mbox{kJ/K}
because \Delta G = \Delta H - T\Delta S where T is constant.

If H changes, then that means that the system is not left to itself but is externally acted upon. So we are not dealing with a spontaneous process.

 

[erty]

If H changes, then that means that the system is not left to itself but is externally acted upon. So we are not dealing with a spontaneous process.

If \Delta H < 0, then enthalpy is released from the system to the surroundings. It would make sense, if I had to add enthalpy (\Delta H > 0, \qquad H = U + pV) and the entropy decreased, but that's not true according to the Gibbs energy.

 

[vivesdn]

If the system is isolated, a spontaneous process will increase its entropy. The Universe is an isolated system in itself, so entropy is always increasing.
But if there is a variation in entalpy (a decrease) at constant temperature T, for sure your system is not isolated. The environment is absorbing such energy, increasing its entropy in a greater extent than the entropy lost by the system (the variation, in absolute value, of the entalpy is grater than that of the entropy, as variation og G is negative). So the universe, the system and the environment, experiments an increase in entropy. This is the result of the second law. No contradiction at all.

 

[vanesch]

If \Delta H < 0, then enthalpy is released from the system to the surroundings. It would make sense, if I had to add enthalpy (\Delta H > 0, \qquad H = U + pV) and the entropy decreased, but that's not true according to the Gibbs energy.

No of course not. Take water. You extract enthalpy (you cool it). It freezes, and its entropy decreases...

 

[erty]

If the system is isolated, a spontaneous process will increase its entropy. The Universe is an isolated system in itself, so entropy is always increasing.
But if there is a variation in entalpy (a decrease) at constant temperature T, for sure your system is not isolated. The environment is absorbing such energy, increasing its entropy in a greater extent than the entropy lost by the system (the variation, in absolute value, of the entalpy is grater than that of the entropy, as variation og G is negative). So the universe, the system and the environment, experiments an increase in entropy. This is the result of the second law. No contradiction at all.

I get it, thanks! \Delta S_{total} > 0, even though \Delta S_{sys} < 0, because \Delta H_{sys} < 0 according to the Gibbs energy and the conditions for a spontaneous process.

Is it possible to calculate the \Delta S_{surroundings}, if I know the \Delta H_{sys}? I mean, is there any proportionality between those two variables?

 

[erty]

No of course not. Take water. You extract enthalpy (you cool it). It freezes, and its entropy decreases...

Yes, but the \Delta S_{total} > 0, because the surroundings get warmer.
(Or did I get this wrong?)

 

[vivesdn]

Is it possible to calculate the \Delta S_{surroundings}, if I know the \Delta H_{sys}? I mean, is there any proportionality between those two variables?

If my memory is still readable, I would say that \Delta S_{surroundings} is equal to \Delta H_{sys} if volume is constant (more generally, if there is no work performed of PV type). Then \Delta H_{sys} is the thermal energy transferred.
If there is work performed, then enthalpy is not equal to Q, the heat.