Why is it wrong to assert that the change in entropy and the change in enthalpy must always have the same sign?

What makes me think that they must have the same sign is the fact that every reaction invariably comes to equilibrium under suitable conditions; and so we have the corresponding temperature equal to $ΔH/ΔS$ (setting $∆G = 0$ in the equation $ΔG = ΔH - TΔS).$

  • 3
    $\begingroup$ Why, it's trivial. Before the reaction comes to equilibrium, it is not in equilibrium, and then $\Delta G\ne0$. $\endgroup$ Jul 23, 2020 at 5:33
  • $\begingroup$ Do you mean the changes in entropy and enthalpy? $\endgroup$ Jul 23, 2020 at 12:04
  • $\begingroup$ @IvanNeretin What I meant is that we can determine (at least theoretically) the temperature at which a reaction comes to equilibrium by using the equation I mentioned in my question. And since temperature is a non-negative quantity, it leads me to think that the change in entropy and the change in enthalpy must have the same sign. $\endgroup$ Jul 23, 2020 at 17:21
  • $\begingroup$ @ChetMiller Yes, my bad. $\endgroup$ Jul 23, 2020 at 17:21
  • $\begingroup$ Temperature is not the only variable affecting $\Delta G$. That being said, yes, when the reaction is in equilibrium, $\Delta H$ and $\Delta S$ must have the same sign. $\endgroup$ Jul 24, 2020 at 6:37

1 Answer 1



In general the entropy of reaction can be written as

$$ T\Delta _r S= \Delta_r H + RT\log \left(\frac{Q_e}{Q}\right) $$

At equilibrium $Q_e=Q$ and

$$ T\Delta _r S_e= \Delta_r H $$

Consider a simple reaction that behaves ideally (occurs under ideal solution conditions).

If it is carried out at constant T and p we can write

$$\Delta_r G = \Delta_r G^\circ + RT \log Q \tag{1}$$

where Q is the reaction quotient.

But we can also write that

$$\Delta_r G = \Delta_r H - T\Delta_r S\tag{2a}$$


$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ\tag{2b}$$

Equation (1) can then be written as

$$\Delta_r G = \Delta_r H^\circ - T(\Delta_r S^\circ-R \log Q) \tag{3}$$

Matching terms in equations (2a) and (3) we have that

$$\Delta_r H = \Delta_r H^\circ \tag{4a}$$


$$\Delta_r S = \Delta_r S^\circ - R\log Q \tag{4b}$$

When the reaction is at equilibrium $Q=Q_e$ (the reaction quotient is then equal to the equilibrium constant, here written $Q_e$) and $\Delta_r G = 0$ which means, combining equations (2a) and (4a) that

$$ T\Delta _r S_e = \Delta _r H^ \circ \tag{5} $$


$$ T\Delta _r S^ \circ = T\Delta_r S_e + RT\log Q_e \tag{6}$$

so that

$$ T\Delta _r S= T\Delta_r S_e + RT\log Q_e - RT\log Q \tag{7a} $$


$$ T\Delta _r S= \Delta_r H ^\circ + RT\log\left(\frac{Q_e}{Q}\right) \tag{7b} $$

Now compare equations (5) and (7b). Equation (5) holds at equilibrium and says, sure enough, that the reaction entropy and enthalpy are equal in sign at this point in the reaction coordinate. However, equation (7b) - which is the more general expression - says that $\Delta_r S$ can in fact differ in sign from $\Delta_r H^\circ$, depending on the magnitude of the reaction quotient Q. It turns out that while the enthalpy of a reaction in an ideal solution is a constant, the entropy of reaction can be tuned by modifying Q.

  • $\begingroup$ Note my notation is not entirely orthodox, but was chosen in part to make the derivations and meanings clear. $\endgroup$
    – Buck Thorn
    Jul 29, 2020 at 20:52
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    $\begingroup$ Thanks a lot for helping me out! @BuckThorn $\endgroup$ Aug 3, 2020 at 18:41

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