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# Entropy

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1. So the question gave us this information:
Entropy is negative
And enthalpy is exothermic

It asked why isn't the reaction feasible at higher temperatures

I jus can't get my head around the answer. Can someone explain? Preferably with examples? Thanks

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2. So the equation for free energy change is ΔG = ΔH - TΔS - as they've given in the mark scheme. Notice you take away entropy change x temperature from enthalpy change. If ΔS is negative you're essentially adding ΔH and TΔS since ΔH - (-TΔS) = ΔH + TΔS. As temperature increases TΔS becomes a more negative number. When the magnitude of TΔS is greater than that of ΔH (which is also negative because the reaction is exothermic) then ΔG is greater than zero at which point the reaction is no longer feasible (remember that for a reaction to be feasible ΔG has to be less than or equal to zero).
3. (Original post by victoria98)
So the equation for free energy change is ΔG = ΔH - TΔS - as they've given in the mark scheme. Notice you take away entropy change x temperature from enthalpy change. If ΔS is negative you're essentially adding ΔH and TΔS since ΔH - (-TΔS) = ΔH + TΔS. As temperature increases TΔS becomes a more negative number. When the magnitude of TΔS is greater than that of ΔH (which is also negative because the reaction is exothermic) then ΔG is greater than zero at which point the reaction is no longer feasible (remember that for a reaction to be feasible ΔG has to be less than or equal to zero).
thanks

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