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∆S (Entropy change) is given by ∆H/T (∆H = Enthalpy change , T= absolute temperature). Then how can the Entropy change for a reaction be negative if the Enthalpy change is positive, and vice versa?

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    $\begingroup$ Who says the enthalpy change is always positive? And what about the equation $\Delta G=\Delta H-T\Delta S$? What if $\Delta G$ is not zero between the initial state of the reactants and the final state of the products? $\endgroup$ – Chet Miller Feb 17 '17 at 12:40
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You are looking at two different entropy changes. The one computed directly from the enthalpy change is the entropy change of the universe. The entropy change of the system is measured by computing the component due to heat transfer. But internal changes in the density of states can also result in a change in entropy, and this can take place without any transfer of heat.

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  • $\begingroup$ What do you mean the same way? You just said that they're two different things, surely they're not measured the same way? $\endgroup$ – RobChem Feb 17 '17 at 11:26
  • $\begingroup$ "Same way" means $dS = dq/T$ @RobChem $\endgroup$ – Zhe Jan 18 '18 at 14:51
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The entropy change you quote is not in general true, but applies only to specific circumstances. Such an instance is the equilibrium between a solid and liquid or between a liquid and its vapour at a temperature T and pressure P.

In Chemistry it is usual to conduct experiments at constant pressure and temperature, rather than at constant volume, and so the place to start with many thermodynamic calculations is the (Gibbs) free energy, $\Delta G=\Delta H - T\Delta S$. As the two phases are in equilibrium the free energy change upon vaporisation is $\Delta G_{vap}=\Delta H_{vap}-T\Delta S_{vap} = 0$ and thus $\Delta S_{vap}=\Delta H_{vap}/T$. Only in the special case when $\Delta G = 0$ does this apply.

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