If a reaction is endothermic and has a confident ΔS, what need to you execute to the temperature in order come make certain the reaction will certainly run?

OK... First, we should relate entropy (ΔS), enthalpy (ΔH, determines even if it is a reaction is endo-/exothermic) and also the spontaneity the the reaction (ΔG, aka Gibbs" free energy)
The equation we space going to need is:ΔG =ΔH - TΔS.... The new symbol (T) is simply temperature in Kelvin.

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A reaction will be voluntary whenΔG is an unfavorable (ΔG = 0, and also will not proceed whenΔG is optimistic (>0)Enthalpy,ΔH, is confident for endothermic reactions and an adverse for exothermic reactions. This may seem counterintuitive, yet you need totake the standpoint the the chemical in this case... If you room a chemical undergoing an exothermic reaction, it way that you are shedding your energy to the surrounding... You are now "energy poorer", for this reason the negative.The temperature, T, is in Kelvin... Therefore it"s always positive.Entropy, ΔS, defines the readjust in the "disorder"... Ok, it"s a simplification, but it functions for our objective here. A optimistic entropy indicates an ext disorder (or much less order).
a reaction that proceeds, which means spontaneous... SoΔG needs to be negative.the reaction to it is in endothermic... SoΔH is positivethe entropy change to it is in positive
(-) = (+) - (+)(+)... Which becomes (-) = (+) - (+) since multiplying two optimistic numbers gives a hopeful number.
So, top top the ideal side the the equation, you have actually a confident number (ΔH) MINUS an additional positive number (TΔS) and you have actually to get a an adverse answer overall.

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This will only take place ifTΔS is bigger thanΔH... And the only way to ensure the this is the case, friend will need to forceTΔS come become big by enhancing the temperature.

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