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### Gibbs Free Energy

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ΔG = ΔH - T ΔSThe Gibbs free energy which is nothing but the change in the free energy of the system and is used as an indicator to tell whether a reaction is spontaneous or not; and whether a reaction is thermodynamically favorable.In this equation: ΔG represents change in Gibbs free energy; ΔH represents change in enthalpy, T represents the temperature in Kelvin, ΔS represents the change in entropy. A reaction is spontaneous if ΔG < 0 or negative.The reaction is non–spontaneous if ΔG > 0 or positive.Now consider the following.For the reaction, the change in enthalpy ΔH = - 437 kJ and change in entropy ΔS = + 272 kJ/ K. For this problem, let’s assume that the value of ΔH and ΔS remains constant at all temperatures.Consider the following reaction.Using this equation ΔG = - 437 – 272*T, the value of ΔG is always negative at all temperatures.For example, if we substitute T = 100 °C (which is = 373.15 K), the equation ΔG becomes –ve as shown here.Similarly at T = -50 °C (which is = 223.15 K), the equation ΔG becomes –ve as shown here.Hence for this reaction, at the given values of ΔH and ΔS, the Gibbs free energy is always negative.This reaction is thermodynamically favorable and spontaneous at all temperatures as long as the values of ΔH are negative and ΔS is positive.Now consider another example which is Haber’s process where Nitrogen gas reacts with hydrogen gas to produce ammonia.For this reaction the value of ΔH =+ 300 kJ and ΔS = + 1000 J/K.The units of ΔS is in joules and this is converted to kilojoules.So, ΔS = 1000 J/K = 1 kJ/K.The value for ΔH and ΔS is substituted into the equation ΔG = ΔH - T ΔS. We obtain an equation ΔG = + 300 – 1* T. The overall sign of ΔG is dependent on the temperature. For example at T = 25 °C which is 298 K, the value of ΔG = + 2 kJ. Since ΔG > 0 (positive sign), the reaction is not thermodynamically favorable.Consider another example where temperature T = 100 °C (which is 373.15 K), the value of ΔG = - 73.15 kJ.The ΔG value is negative that is less than 0. So, the reaction is thermodynamically favorable. Hence, for the given value of enthalpy and entropy, ΔG < 0 only at high temperatures.The table here represents the conditions at which a reaction becomes thermodynamically favorable.This is a shortcut method used to determine thermodynamic favorability depending on the sign associated with ΔH and ΔS.