Gibbs Free Energy Evaluating spontaneity
Predicting Spontaneity An increase in entropy; Changing from a more structured to less structured physical state: Solid to liquid Liquid to gas Increase in temperature causing a breakdown of structure within the states Fewer molecules becoming more molecules Exothermic; A lower energy state releasing energy is favorable
Gibbs Free Energy Equation ΔG = ΔH - TΔS If ΔG is negative (< 0) the reaction is favorable or spontaneous. If ΔG is positive (> 0) the reaction is unfavorable or nonspontaneous ΔH is the enthalpy change of the system T is the temperature at which the reaction is occurring (in kelvin) ΔS is the entropy change of the system
Predicting Spontaneity based on the equation ΔG = ΔH - TΔS ΔS ΔH Exothermic (-) Endothermic (+) Increasing Entropy (+) Decreasing Entropy (-) Always Spontaneous at any temperature Only spontaneous at low temperatures Only spontaneous at high temperatures Never Spontaneous (but reverse is)
Enthalpy and Spontaneity Problem 2NI 3 (s) N 2 (g) + 3I 2 (g) Is ΔH for the decomposition of NI 3 positive or negative? ΔH = -384 kj/mol Is ΔS positive or negative? Definitely positive, as it is 2 moles becoming 4, and it is a solid turning into gas. ΔH ΔS Exothermic (-) Endothermic (+) Increasing Entropy (+) Always Spontaneous at any temperature Only spontaneous at high temperatures Decreasing Entropy (-) Only spontaneous at low temperatures Never Spontaneous (but reverse is)
Distinction Between Spontaneity and Equilibrium A + B AB Equilibrium shifts gives the percentage of products and reactants under certain conditions. But, implies that it is still spontaneous and some of each side are present. If ΔG is > 0 (positive), then there is no equilibrium and NO products are present.
Reversible Reactions, Entropy, and Le Chatelier s Principle 2NO 2 (g) N 2 O 4 (g) NO 2 is a brown gas, while N 2 O 4 is a clear gas Which direction is favored at different temperatures? ΔG = ΔH - TΔS
Reversible Reactions, Entropy, and Le Chatelier s Principle 2NO 2 (g) N 2 O 4 (g) Is the ΔS for this reaction positive or negative? ΔS should be negative because it is becoming more orderly. ΔH f of NO 2 = 33.18 kj/mol ΔH f of N 2 O 4 = 9.16 kj/mol Is this reaction exothermic or endothermic?
Reversible Reactions, Entropy, and Le Chatelier s Principle 2NO 2 (g) N 2 O 4 (g) Is the ΔS for this reaction positive or negative? ΔS of NO 2 = 240.06 J/mol*K ΔS of N 2 O 4 = 304.24 J/mol*K Is this reaction exothermic or endothermic? -57.2 kj/mol means it is exothermic
Reversible Reactions, Entropy, and Le Chatelier s Principle 2NO 2 (g) N 2 O 4 (g) Is the ΔS for this reaction positive or negative? -175.38 J/mol*K It is indeed more negative Is this reaction exothermic or endothermic? -57.2 kj/mol means it is exothermic
Is it Spontaneous with ΔS at -175.38 and ΔH at -57.2? ΔG = ΔH - TΔS ΔS ΔH Exothermic (-) Endothermic (+) Increasing Entropy (+) Decreasing Entropy (-) Always Spontaneous at any temperature Only spontaneous at low temperatures Only spontaneous at high temperatures Never Spontaneous (but reverse is)
Is it Spontaneous with ΔS at -175.38 and ΔH at -57.2? ΔG = ΔH TΔS 2NO 2 (g) N 2 O 4 (g) ΔS ΔH Exothermic (-) Endothermic (+) Increasing Entropy (+) Decreasing Entropy (-) Always Spontaneous at any temperature Only spontaneous at low temperatures Only spontaneous at high temperatures Never Spontaneous (but reverse is)
Is it Spontaneous with ΔS at -175.38 and ΔH at -57.2? ΔG = ΔH TΔS 2NO 2 (g) N 2 O 4 (g) Spontaneous at low temperatures means at lower temperatures, more N 2 O 4 will be created at the color should turn more clear. At higher temperatures, more should remain NO 2 and be brown.
Reversible Reactions, Entropy, and Le Chatelier s Principle 2NO 2 (g) N 2 O 4 (g) Is the ΔS for this reaction positive or negative? -175.38 J/mol*K Is this reaction exothermic or endothermic? -57.2 kj/mol means it is exothermic Is this spontaneous at 25 C using ΔG = ΔH TΔS? ΔG = -4.94 so yes, it is spontaneous
Le Chatelier s Principal and Temperature 2NO 2 (g) N 2 O 4 (g) ΔH = -57.2 kj/mol Evaluating the effects of temperature on equilibrium shifts are based on whether the forward direction is endothermic or exothermic. If the reaction is exothermic, an increase in temperature shifts the reaction left. If the reaction is endothermic, an increase in temperature shifts the reaction right.
Reversible Reactions, Entropy, and Le Chatelier s Principle 2NO 2 (g) N 2 O 4 (g) NO 2 is a brown gas, while N 2 O 4 is a clear gas
Le Chatelier s Principal and Temperature 2NO 2 (g) N 2 O 4 (g) If the reaction is exothermic, an increase in temperature shifts the reaction left. If the reaction is endothermic, an increase in temperature shifts the reaction right. Let s Discuss Why...
Le Chatelier s Principal and Temperature 2NO 2 (g) N 2 O 4 (g) ΔG = ΔH TΔS 0 = -57.2kJ/mol T(-0.17538 kj/mol*k) 57.2 = 0.17538T T = 325 K or 53 C If you have a temperature lower than 326K, the ΔG is negative, and the reaction is spontaneous. If you have a temperature above that, the reaction is not spontaneous andn can not proceed, thus only NO 2 will be present.