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Gibbs Free Energy Change

Gibbs Free Energy Change. Spontaneous Reactions in Nature Reactions in nature are spontaneous based on two factors Change in enthalpy Change in entropy All reactions in nature generally want to lose heat And increase disorder

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Gibbs Free Energy Change

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  1. Gibbs Free Energy Change • Spontaneous Reactions in Nature • Reactions in nature are spontaneous based on two factors • Change in enthalpy • Change in entropy • All reactions in nature generally want to lose heat • And increase disorder • But some reactions are spontaneous even if they don’t follow this pattern. • Freezing ice • Lowers heat • loses randomness • Boiling water • Gains heat • Gains randomness • Temperature plays an important role in spontaneous reactions. H S -H +S Spontaneous at low temperature -H -S +H +S Spontaneous at high temperature

  2. II. Gibbs Free Energy Change A. Defined ΔGo = ΔHo – T ΔSo ΔH = Change in enthalpy ΔS = change in entropy T = absolute temp ΔG = Gibbs free energy (free energy) B. Meaning If ΔG is negative If ΔG is positive If ΔG is zero C. Predicting ΔG Since temperature is always positive, we can predict if a reaction is spontaneous by following some general rules kJ/mol J/mol  K K kJ/mol Reaction occurs spontaneously Reaction is not spontaneous, reverse reaction is spontaneous Reaction is at EQUILIBRIUM

  3. - + - Always spontaneous - + - + + Never spontaneous Only spontaneous if temp is low (keeps –TΔS from getting too big) - - + Only spontaneous if temp is high (makes TΔS bigger than ΔH) + - + Example – When will the reaction A(s) + B(s) C(g) + 4 kJ be spontaneous? Since we are making more gas molecules, ΔS is positive Since heat is release, ΔH is negative This is always spontaneous

  4. Example – What will happen to the spontaneity of the reaction 2 H2(g) + O2(g)  2 H2O(g) + 49.2 kJ as temperature increases? Since the ΔS of the reaction is negative (fewer gas molecules) and the ΔH of the reaction is negative, this will become less spontaneous as temperature increases. (if water is heated too much, it will spontaneous decompose into hydrogen and oxygen gas!) D. Calculating ΔGo 1.From standard ΔG values Like ΔS and ΔH, we can use the standard free energy values to calculate ΔG. ΔGo = ΣnGoprod -ΣmGoreact Like Hfo, the Go value of any substance in its standard state is zero.

  5. Example- Using the table in your textbook, determine if the following reactions are spontaneous, based on the free energy change. a. N2O4 2 NO2 ΔG = 2(GoNO2) – (GoN2O4) ΔG = 2(51.84) – 98.28 ΔG = 5.4 kJ/mol b. Al2O3 + 3 H2(g) 2 Al(s) + 3 H2O(g) ΔG = (3(-228.57) + (2(0)) – (-1576.5 + 3(0) ΔG = 890.79 kJ/mol Not spontaneous at 25oC!

  6. E. Calculating ΔG from ΔH and ΔS You can calculate the ΔH and ΔS for a reaction by using the standard tables, then use the temperature to determine the free energy change. You must convert ΔS values to kJ!! Example – Using your tables, show that ΔGo = ΔHo – T ΔSo for the reaction 2 H2(g) + O2(g) 2 H2O(g) at 25oC. You need to calculate ΔS and ΔH for the reaction, then use the free energy formula Entropy ΔS = 2(188.83) –(2(130.58)+205) ΔS = -88.5 J/mol K ΔS = -0.0885 kJ/mol K Enthalpy H = 2(-241.82) –(2(0)+0) H = -483.64 kJ/mol This is exactly 2 times the free energy change for one mole of water! ΔGo = ΔHo – T ΔSo =-483.64 –(298)(-0.0885) = -457.14

  7. Example – Given the ΔH = -61.0 kJ/mol and ΔS= -0.052 kJ/mol K for the reaction A(g) + B(s) C(g) , at what temperature range this reaction will be spontaneous. Since S and H are negative, this is spontaneous at low temperatures. Calculate the temperature needed to get G=0, Any temp below this will make ΔG negative ΔG = ΔH – TΔS 0 = -61 –(x)(-0.052) -61 = x(-0.052) X = 1173 K 900oC Any temperature below this will make ΔG negative (spontaneous)

  8. III. Using Free Energy equation when not at standard conditions. A. Finding ΔG If we are not at standard conditions, we can calculate ΔG by using ΔG = ΔGo + RT ln(Q) ΔGo= T= Q= R= Standard free energy Temp in Kelvin Reaction quotient (calculated same as Keq) Other gas constant 8.314 J/mol K When using this equation, use the R value that has the same units as entropy! Under standard conditions, all concentrations are 1.0 M, so Q would be 1, and ln(Q) would be zero, so RT ln(Q) drops off.

  9. Example – Given the ΔGo = -61.0 kJ/mol for the reaction A(g) + B(s)  C(g), what is the free energy change if we have 2.0 moles of A, 3 moles of B and 5 moles of C in a 1.0 L container, and the temperature is increased to 30.0oC? First we need to calculate Q, then we can use the equation ΔG = ΔGo + RT ln(Q) Q = [C] [A] Q = 5/2 Q = 2.5 ΔG = ΔGo + RT ln(Q) ΔG = -61 + (0.008314 kJ/mol k)(303)ln(2.5) ΔG = -58.69 kJ/mol

  10. B. Calculating Go at equilibrium If we are at equilibrium, there are two important points to remember ΔG = 0 This changes our formula to ΔGo = -RT ln (Keq) If we know the Keq, we can determine the standard free energy change at any temperature. Q = Keq

  11. Example – Using the standard Go values, calculate the equilibrium constant of the following reaction at 50.0oC. • CO (g) + ½ O2 (g) CO2(g) • Calculate ΔGo for this reaction • GCO2 – (GCO + ½ (GO2)) = ΔG • -394.4 kJ – (-137.2 kJ + ½ (0)) • ΔG = -257.2 kJ/mol • 2. Use ΔG = -R T ln (Keq) • -257.2 = - 0.008314 (323K) ln Keq • 95.7 = ln Keq • Keq = 3.65 x 1041

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