If a reaction is endothermic and has a positive ΔS, what should you do to the temperature in order to make sure the reaction will run?
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OK... First, we need to relate entropy (ΔS), enthalpy (ΔH, determines whether a reaction is endo-/exothermic) and the spontaneity of the reaction (ΔG, aka Gibbs" Free energy)
The equation we are going to need is:ΔG =ΔH - TΔS.... The new symbol (T) is simply temperature in Kelvin.

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A reaction will be spontaneous whenΔG is negative (ΔG = 0, and will not proceed whenΔG is positive (>0)Enthalpy,ΔH, is positive for endothermic reactions and negative for exothermic reactions. This may seem counterintuitive, but you need totake the standpoint of the chemical in this case... if you are a chemical undergoing an exothermic reaction, it means that you are losing your energy to the surrounding... you are now "energy poorer", hence the negative.The temperature, T, is in Kelvin... so it"s always positive.Entropy, ΔS, describes the change in the "disorder"... ok, it"s a simplification, but it works for our purpose here. A positive entropy indicates more disorder (or less order).
a reaction that proceeds, which means spontaneous... soΔG has to be negative.the reaction to be endothermic... soΔH is positivethe entropy change to be positive
(-) = (+) - (+)(+)... which becomes (-) = (+) - (+) since multiplying two positive numbers gives a positive number.
So, on the right side of the equation, you have a positive number (ΔH) MINUS another positive number (TΔS) and you have to get a negative answer overall.

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This will only happen ifTΔS is larger thanΔH... and the only way to ensure that this is the case, you will need to forceTΔS to become large by increasing the temperature.
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