Chemical Equilibrium

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The fundamentals of chemical equilibrium including Le Chatier's Principle and solved problems for heterogeneous and homogeneous equilibrium.
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Chemical Equilibrium

  1. 1. Copyright Sautter 2015
  2. 2. EQUILIBRIUM • In many chemical reactions both a forward and reverse reaction occur simultaneously. • When the rate of forward and reverse reactions are equalized, the system is at equilibrium. • Not all reactions can reach equilibrium. Consider a burning log. The products of the combustion (the smoke and ashes) can never reunite to form the reactants (the log and the oxygen) • By contrast some common processes are equilibrium systems, such as the formation of a saturated solution or the vaporization of a liquid in a sealed container. 2
  3. 3. RECOGNIZING EQUILIBRIUM • When equilibrium systems do exist they may be recognized by their apparent static nature (it looks as if nothing is happening). Additionally, all equilibrium systems must be closed, that is, nothing let in and nothing let out including energy. • While equilibrium appear unchanging when observed on a large scale (macroscopically) they are in constant change on the molecular level (microscopically). Reactants are continually forming products and products are continually forming reactants at equal rates. • Simply stated, equilibrium systems are macroscopically static and microscopically dynamic(changing). 3
  4. 4. SOLUBILITY EQUILIBRIUM RATE OF DISSOLVING = RATE OF CRYSTALIZATION RATE OF FORWARD PROCESS = RATE OF REVERSE PROCESS The amount of substance dissolved remains constant The system appears macroscopically static. 4
  5. 5. CHEMICAL EQUILIBRIUM • When the rates of opposing processes are equal, equilibrium has been established whether the system is physical or chemical. • Given the reaction: a A + b B  c C + d D when the rate at which A reacts with B equals the rate at which C reacts with D the system is at equilibrium. • Recall that the rate equation for a reaction is: Rate = k x [reactants]n • At equilibrium Rate forward = Rate reverse or: kf [reactants] = kr[products] and rearranging the equation: kf / kr = [products] / [reactants] and a constant divided by a constant gives another constant so we get: • Ke = [PRODUCTS] / [REACTANTS] at equilibrium for the given reaction then is: • Ke = ([C]c x [D]d) / ([A]a x [B]b) notice that in this equation called the equilibrium expression, the coefficients of the balanced equation serve as powers. 5
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  7. 7. THE EQUILIBRIUM EXPRESSION • For a system at equilibrium, the value of the equilibrium constant (Ke) remains constant unless the temperature is changed. The concentrations of products and reactants may change as a result of “equilibrium shifts” but the ratio of products to reactants as calculated using the equilibrium expression remains unchanged. • Let’s find the equilibrium expression for the following reaction: 2 H2(g) + O2(g)  2 H2O(g) (the double arrow  means equilibrium) • Ke = [H2O] 2 / ([H2]2 x [O2]) • How about H2(g) + 1/2 O2(g)   H2O(g) ? The reaction is the same but it is balanced with a different set of coefficients and therefore the equilibrium expression is different: Ke = [H2O] / ([H2] x [O2 ]1/2) 7
  8. 8. THE EQUILIBRIUM EXPRESSION • The coefficients used in balancing the reaction alters the equilibrium expression and also the value of the equilibrium constant. • For example, using the equilibrium equation H2(g) + I2(g)  2 HI(g) the equilibrium expression is Ke = [HI] 2 / ([H2] x [I2]) Let us assume the following equilibrium concentrations [H2] = 0.094 M, [I2] = 0.094 and [HI] = 0.012 M Ke = (0.012)2 / ((0.094) x (0.094)) = 0.016 • If the equation was balanced as ½ H2(g) + ½ I2(g)  HI(g) the equilibrium expression would be: Ke = [HI] / ([H2]1/2 x [I2] 1/2 ) and Ke = (0.012) / ((0.094)1/2 x (0.094)1/2 ) = 0.127 • The numerical value of the equilibrium constant depends on the coefficients used to balance the equation! 8
  9. 9. HOW ARE Ke VALUES RELATED? • In our previous example when the coefficients of the balanced equation were halved the Ke value changed. But how? • H2(g) + I2(g)  2 HI(g) , Ke = [HI] 2 / ([H2] x [I2]) = 0.016 • ½ H2(g) + ½ I2(g)  HI(g) , Ke = [HI] / ([H2]1/2 x [I2] 1/2 ) = 0.127 • Notice that the square root (half power) of 0.016 = 0.127 • If for some reason the equation was balanced as: 2 H2(g) + 2 I2(g)  4 HI(g), the constant would change to the square of the original value or 0.0162 and Ke would equal 0.064 • When the balancing coefficients of an equation are changed by a factor, the Ke value changes by that factor as a power of the original Ke 9
  10. 10. HOW ARE Ke VALUES RELATED? • How else may the original balanced equation be altered? • Often equations are reversed (interchanging products with reactants. For example: H2(g) + I2(g)  2 HI(g) may be rewritten as: 2 HI(g)  H2(g) + I2(g) The equilibrium expression for the first equation is: Ke = [HI] 2 / ([H2] x [I2]) , for the second it is: Ke = ([H2] x [I2]) / [HI] 2 , the reciprocal of the first • Therefore when an equation is reversed, the Ke becomes the reciprocal of the original Ke value. • Ke (reverse) = 1 / K (forward) 10
  11. 11. HETEROGENEOUS EQUILIBRIUM SYSTEMS • Heterogeneous and homogeneous are terms that are applied to the physical state (phases) of the components in equilibrium systems. A system consisting of all gases is heterogeneous while one including, for example, both solids and gases is heterogeneous. • H2(g) + I2(g)  2 HI(g) (homogeneous – all gases) • H2(g) + ½ O2(g)  H2O(l) (heterogeneous – gas & liquid) • Solids and liquids have concentrations which can vary only slightly due to temperature changes. For all practical proposes the concentrations of solids and liquids are fixed. Only gases and dissolved substances can change concentration. Since the equilibrium expression involves only components which change concentration, solids and liquids are not included in its format. 11
  12. 12. HETEROGENEOUS EQUILIBRIUM SYSTEMS • When solids or liquids are encountered in an equilibrium equation, a “1” is substituted in the equilibrium expression for that component. • 2 H2(g) + O2(g)  2 H2O(l) Ke = 1 / ([H2]2 x [O2]) • For a system such as: CaCO3(s)  CaO(s) + CO2(g) The equilibrium expression is: Ke = 1 x [CO2] / 1 or simply Ke = [CO2] notice that since both CaCO3(s) and CaO(s) are solids they have been replaced by ones in the equilibrium expression. 12
  13. 13. THE MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS Equilibrium means that the rate of the forward reaction and the rate of the reverse reaction are equal. The concentrations of the products and the reactants need not be equal and rarely are equal at equilibrium When the equilibrium constant is small, the concentrations of the reactants are large as compared to the products at equilibrium. (The equilibrium favors the reactants). 13
  14. 14. THE MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS When the equilibrium constant is large, the concentrations of the reactants are small as compared to the products at equilibrium. (The equilibrium favors the products). 14
  15. 15. THE MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS When the equilibrium constant is about one, the Concentrations of the reactants and the products are about equal at equilibrium. 15
  16. 16. MEASURING Ke VALUES • Since concentration may be measured for solutions in moles per liter and for gases in terms of pressures (atms) or moles per liters two unit systems are possible for Ke. • When the concentration values used in calculating Ke are in terms of molar units the constant is referred to as Kc (for concentration units – moles per liter). When concentration of gases is measured in pressure units (atms) the constant is referred to as Kp (for pressure units – atms). • The equilibrium constant in concentration terms (Kc) can be converted to an equivalent value in pressure units (Kp). • Kp = Kc(R x T) n , in this equation: R = 0.0821 atm l / moles K, T = Kelvin temperature n = moles of gaseous products – moles of gaseous reactants in the balanced equation. 16
  17. 17. CONVERTING KP AND KC VALUES • Problem: For the reaction: N2(g) + 3 H2(g)  2 NH3(g) , Kc = 0.105 at 472 0C. Find Kp at this temperature. • Solution: Kp = Kc(R x T) n , R = 0.0821 atm x l / mole x K T= 472 + 273 = 745 K, n = 2 – 4 = - 2 (two moles of NH3(g) product gases) – (one mole of N2(g) and three moles of H2(g) a total of four moles of reactant gases) • Kp = 0.105 x (0.0821 x 745)-2 = 2.81 x 10-5 17
  18. 18. CALCULATING THE EQUILIBRIUM CONSTANT • Problem: If 0.20 moles of PCl3(g) and 0.10 moles of Cl2(g) are placed in a 1.0 liter container at 200 0C the equilibrium concentration of PCl5 is found to be 0.012 molar. What is Kc for the reaction: PCl3(g) + Cl2(g)  PCl5(g) • Solution: Since one PCl3 and one Cl2 must be consumed to form one PCl5, (0.20 M, the starting molarity of PCl3 – 0.012 M, the molarity of PCl5 formed) = 0.188 M PCl3 remains and (0.10 M, the starting molarity of Cl2 – 0.012 M the molarity of PCl5 formed) = 0.088 M Cl2 remaining.. • Kc = [PCl5] / ([PCl3] x [Cl2]) • Kc = (0.012) / (0.188 x 0.088) = 0.725 • NOTE: The temperature given in the problem is not used in the calculation. It is refer data only. Temperatures are not used in equilibrium calculations of this type! 18
  19. 19. USING THE EQUILIBRIUM CONSTANT • Problem: For the reaction: H2(g) + I2(g)  2 HI(g) , Kc = 0.016 at 793 K. If 1.00 mole of hydrogen and 1.00 mole of iodine are mixed in a 10.0 liter container, what is the equilibrium concentration of all components? • Solution: H2(g) + I2(g)  2 HI(g) and therefore, Ke = [HI] 2 / ([H2] x [I2]) The starting [H2] and [I2] both equal 1.00 mole / 10.0 liters or 0.100 M. • The concentration of HI at equilibrium is unknown (X). • In order to form 2 HI, only 1 H2 and 1 I2 (half as many of each) are needed. The equilibrium [H2] and [I2] are both then (0.10 - .5X). This represents the starting concentrations of each minus the concentration used to form the HI. 19
  20. 20. USING THE EQUILIBRIUM CONSTANT (cont’d) • Substituting the equilibrium concentrations into the equilibrium expression we get: • Kc = X2 / (0.10 – 0.5 X)(0.10 – 0.5X) = 0.016 or • X2 / (0.10 – 0.5X)2 = 0.016, taking the square root of both sides of the equation we get: • X / (0.10 – 0.5X) = 0.126 and X = 0.0126 – 0.008X • Rearranging the equation to solve for X we get: • 1.008X = 0.0126, X = 0.0126 / 1.008 = 0.0125 • [HI]e = 0.0125 M, • [H2]e = [I2]e = 0.10 – .5(0.0125) = 0.0938 M • CHECK – Placing the equilibrium concentrations back into the equilibrium expression should give the correct constant! • (0.0125)2 / (0.938)2 = 0.017 a very close approximation of Kc (0.016) considering the rounding errors! 20
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