At constant pressure, which is the usual condition for chemical reactions, heat absorbed/released by the system is the same thing as enthalpy change. According to the Gibbs equation, $$\mathrm dG = \mathrm dH-T\mathrm dS = \mathrm dH-\mathrm dQ$$ But, $\mathrm dH=\mathrm dQ$ at constant a pressure. So is Gibbs energy change zero for any reaction?

Another problem I face is the explanation in my book. It takes many cases, for endothermic, exothermic etc. One of the cases is: Exothermic process: $\mathrm dH$ negative, $\mathrm dS$ positive, spontaneous,

But, for exothermic, isn’t $\mathrm dQ$ always negative? So how can $\mathrm dS$ be positive?

  • $\begingroup$ possible duplicate of Gibbs free energy-minimum or zero? $\endgroup$ – user15489 Sep 12 '15 at 3:32
  • $\begingroup$ No, I am not talking about equilibrium at all, just the basic definition of Gibbs free energy change, poses a confusion at constant pressure $\endgroup$ – Newton Sep 12 '15 at 3:41
  • $\begingroup$ Understanding Gibbs free energy $\endgroup$ – user15489 Sep 12 '15 at 3:43
  • $\begingroup$ Doesn't seem to answer my question, according to which change in free energy is zero for any reaction at constant pressure. $\endgroup$ – Newton Sep 12 '15 at 4:01

I have issues with the other answer, which seems to mix finite quantities with infinitesimal quantities, and does not even address the main crux of the issue which is the reversibility of the process. Instead, it only tries to explain the physical significance of the terms $\Delta H$ and $-T\Delta S$ in a vague way, so I am not even really sure how it answers your main question. Even if this does not help you directly, it may be of use to others who read this question in the future.

You wrote that:

$$\mathrm{d}G = \mathrm{d}H - T\mathrm{d}S = \mathrm{d}H - đq$$

The issue with the second equality is that it assumes that the process is reversible. Without going into too much detail (it can be found easily elsewhere), entropy is defined by the following equation:

$$\mathrm{d}S = \frac{đq_\mathrm{rev}}{T}$$

where the subscript $\mathrm{rev}$ indicates a reversible process. It can be shown that for an irreversible process,

$$\frac{đq_\mathrm{irrev}}{T} < \frac{đq_\mathrm{rev}}{T} = \mathrm{d}S$$

This result is the Second Law of Thermodynamics. Therefore, for an irreversible process, $T\mathrm{d}S > đq$.

You are right in saying that at constant pressure, $\mathrm{d}H = đq$. Putting it all together, we have:

$$\mathrm{d}G = \mathrm{d}H - T\mathrm{d}S = đq - T\mathrm{d}S < 0 \qquad \text{(irrev. process; closed system; }p, T\text{ const.)}$$

Therefore, the direct answer to the question is: $\mathrm{d}G$ is, in general, not equal to $0$ because $T\mathrm{d}S \neq đq$.

In fact, you may notice that the condition $\mathrm{d}G < 0$ is the condition for spontaneity under constant $p$ and $T$. This is no coincidence. The concepts of spontaneity (in the context of $G$) and irreversibility (in the context of $S$) are extremely similar, in that they both refer to a tendency for a process to proceed.

For a reversible process, $T\mathrm{d}S = đq$. Following the logic that you used in your question, this means that $\mathrm{d}G = 0$, which is correct. If $\mathrm{d}G = 0$, the forward process cannot be spontaneous (that would require $\mathrm{d}G < 0$) and the reverse process cannot be spontaneous (that would require $\mathrm{d}G > 0$).

This can only mean one thing: the system is in equilibrium. Indeed, the condition for equilibrium at constant $p$ and $T$ is $\mathrm{d}G = 0$. And at this point, it should not come as a surprise that a reversible process is, by definition, one that is at equilibrium throughout.

  • 1
    $\begingroup$ Your answer is the best $\endgroup$ – Newton Oct 14 '15 at 16:53

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And, all exothermic reactions are not spontaneous. Spontaneity of any system is not individually esplained by enthalpy or entropy. It is only by ΔG (-ve for spontaneous and +ve for non-spontaneous). For expressing ΔG value, enthalpy and entropy with respect to Temperature could be considered. Thank u.

  • $\begingroup$ The answer was in the basic definition itself, but I couldn't see it. $\endgroup$ – Newton Sep 12 '15 at 5:43

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