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Equilibrium Chemical Equilibrium. General Info on Equilibrium Concerned with how far a reaction goes. Why does it have a low or high % yield? Why do some.

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Presentation on theme: "Equilibrium Chemical Equilibrium. General Info on Equilibrium Concerned with how far a reaction goes. Why does it have a low or high % yield? Why do some."— Presentation transcript:

1 Equilibrium Chemical Equilibrium

2 General Info on Equilibrium Concerned with how far a reaction goes. Why does it have a low or high % yield? Why do some of the reactants never become products? Not concerned w/ how fast (not looking at rates) At equilibrium, a reaction: –Macroscopically looks finished –Microscopically the forward reaction rate = the reverse reaction rate –Has both reactants and products present but not necessarily in equal concentrations or amounts. –Is in a closed system

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4 What determines where equilibrium is reached? (Why do some reactions have low yields and others high yields?) The equilibrium state is a compromise between a loss in enthalpy ( in H) and a gain in entropy ( in S). –S = change in entropy. Entropy is a measurement of chaos. H 2 0 (l) H 2 O (g) the forward reaction has a gain in S and a gain in H. = equilibrium Reactions which do not have this compromise such as combustion reactions, have no measurable equilibrium. The reaction only goes forward. Ex. CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (g) this reaction loses enthalpy and no change in entropy when going in the forward direction. = no equilibrium

5 Quantitative Equilibrium The equilibrium math expression: – K = [Products]/[Reactants] K= the equilibrium constant –Rules for writing the K expression: i) do not include solids or liquid water ii) Coefficients are written as exponents iii) Substitute into the expression only equilibrium concentrations Kc or equilibrium pressures Kp.

6 K constant The K constant for a reaction stays the same unless temperature is changed. (Changing temp. will change the K constant because it changes the thermodynamics of the reaction system. A new compromise between S and H is reached) Larger K constant = greater concentration of the products at equilibrium. Therefore, there is a higher % yield, or the forward reaction is more favored.

7 K constant cont. Categories of K constants and expressions: K a = Acids K b =Bases K p =gases using partial pressures K c = using molar concentrations

8 Le Châteliers principle Any change forced upon a reaction system at equilibrium will cause the reaction to respond in such a way as to counter act that change and restore equilibrium. Concentrations effects: –A change in concentration only changes the position of equilibrium (shifts equilibrium) but does not change the K constant.

9 Le Châteliers principle cont. Concentrations effects (cont.) An increase of the concentration of a reactant or a decrease in the concentration of the product causes the reaction to shift in the forward direction to produce a product. A decrease in the concentration of a reactant or an increase in the concentration of the product causes the reaction to shift in the reverse direction.

10 Le châteliers principle cont. Pressure (similar to concentration) does not effect the K constant. It only affects gases The side (reactants or products) which has the most moles of gas will experience the effect of the pressure change to the greatest extent. An increase in pressure = an increase in concentration. A decrease in pressure = a decrease in concentration. An increase in pressure causes the reaction to shift in the forward direction if there are more moles of gases as reactants A decrease in pressure causes the reaction to shift in the reverse direction with more moles of gases as reactants

11 Le Châteliers principle cont. Temperature: alters both the position and the constant –Endothermic: an increase in temperature causes a shift in the forward direction and an increase in the K constant, because there will be more products than reactants present –Exothermic: an increase in temperature causes a shift in the reverse direction and a decrease in the K constant. Less products present at equilibrium.

12 Le châteliers principle cont. Catalyst: has no effect on the position or the constant

13 Haber Process This reaction is used for the mass production of ammonia The reaction is: N 2 + 3H 2 2NH 3 all (g) ΔH = -92kJ K= [NH 3 ] 2 / [N 2 ] [H 2 ] 3

14 The Contact Process

15 The industrial process of making ammonia

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17 Yields of NH 3 at various pressures and temperatures

18 Summary of the Haber process Is a great example for the application of Le Chateliers principle Increases the pressure (300atm) to create a forward shift. Constantly supplies the reactants H 2 & N 2 Removes the NH 3 through cooling (NH 3 has hydrogen bonding)

19 Also applies principles of kinetics. –Uses a Fe catalyst which lowers the E a –Uses high temperatures to speed up the reaction rate. (The reaction will not reverse since there is no ammonia available)

20 The Contact Process This process consists of a series of reactions which produce sulfuric acid. S (s) + O 2(g) SO 2(g) 2SO 2(g) + O 2(g) 2SO 3(g) ΔH -196 kJ/mol SO 3(g) + H 2 O (l) H 2 SO 4(l)


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