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For an endothermic reaction with a negative change in entropy, based on dG = dH - TdS increasing the temperature should make the process "less spontaneous". However, if the reaction is endothermic then classic Le Chatelier principle says that the reaction should shift right with increasing temperature. How can we reconcile this?

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  • $\begingroup$ Are you familiar with the Van't Hopf Equation and with its derivation? $\endgroup$ – Chet Miller Mar 7 '18 at 20:56
  • $\begingroup$ Hi Chester, yes I am. How does this help? $\endgroup$ – Andrew U Mar 8 '18 at 7:38
  • $\begingroup$ The derivation explains the correct way to take into account the effect of temperature on the equilibrium constant. The predictions from Van't Hopf are consistent with Le Chatelier. Let's see your analysis using Van't Hopf. $\endgroup$ – Chet Miller Mar 8 '18 at 12:27
  • $\begingroup$ dGo = dHo – TdSo = -RT lnK Rearranging the second half of the equation gives (–dHo/RT) + (dSo/R) = lnK So we have a linear relationship for lnK vs 1/T, with slope –dHo/R $\endgroup$ – Andrew U Mar 8 '18 at 22:33
  • $\begingroup$ So, from this equation, what is the sign of the derivative of lnK with respect to temperature if the reaction is exothermic? $\endgroup$ – Chet Miller Mar 9 '18 at 1:36
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The change in free energy $\Delta G^0$ in these equations represents the change from molar quantities of reactants (in separate containers) at temperature T and pressure 1 bar, and corresponding molar quantities of products (in separate containers) at temperature T and pressure 1 bar. So writing the equation in terms of differentials makes no sense. It should more properly represented by $$\Delta G^0=H^0_{P}-H^0_R-T(S^0_P-S^0_R)$$where the subscript P represents products and the subscript R represents reactants. So, $$\frac{\Delta G^0}{T}=\frac{H^0_{P}-H^0_R}{T}-(S^0_P-S^0_R)$$Now, for $H^0_P$ and $H^0_R$, since the pressure is constant between the two states of the products and reactants, respectively, at different temperatures, we have $$dH^0_P=(C_p)_PdT=TdS^0_P$$and $$dH^0_R=(C_p)_RdT=TdS^0_R$$So, algebraically, by the product rule for differentiation, we have$$\frac{d(\Delta G^0/T)}{dT}=-\frac{(H_P^0-H_R^0)}{T^2}+\frac{T(dS_P^0-dS_R^0)}{TdT}-\frac{(dS^0_P-dS^0_R)}{dT}=-\frac{(H_P^0-H_R^0)}{T^2}$$This is what you get when you do the math correctly.

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  • $\begingroup$ My differentiation is a little rusty so I didn't follow the very last step, but essentially entropy out of the temperature dependence, correct? $\endgroup$ – Andrew U Mar 22 '18 at 22:12
  • $\begingroup$ Also, just to clarify, when I wrote dGo = dHo - T dSo I was not referring to differential quantities, rather d for delta. $\endgroup$ – Andrew U Mar 22 '18 at 22:13
  • $\begingroup$ That d and delta representation certainly confused me. With regard to your previous comment, the differential of uv is udv+vdu. What this leads to is that one of the terms from differentiating the enthalpy term with respect to temperature cancels with the temperature derivative of the entropy term. $\endgroup$ – Chet Miller Mar 22 '18 at 22:18
  • $\begingroup$ So if we go back to the earlier discussion, because (delta)H does indeed depend on Temp (which we typically ignore), is it too simplistic to simply say that for an endothermic reaction with negative entropy change, increasing the temperature makes the reaction "less spontaneous" because we cannot simply plug into (delta)G = (delta)H - T (delta)S? $\endgroup$ – Andrew U Mar 23 '18 at 17:19
  • $\begingroup$ Yes. The enthalpy dependence with temperature cancels with the entropy dependence with temperature. $\endgroup$ – Chet Miller Mar 23 '18 at 17:49

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