Chemical Equilibrium
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Chemical Equilibrium

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

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

    • CHEMICAL EQUILIBRIUM Copyright Sautter 2003
    • The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www.wsautter.com
    • Books available at: www. wsautter .com www.smashwords.com www.amazon.com www.bibliotastic.com www.goodreads.com Walt’s Books for Free!
    • 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.
    • 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).
    • 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.
    • 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:
      • k f [reactants] = k r [products] and rearranging the equation:
      • k f / k r = [products] / [reactants] and a constant divided by a constant gives another constant so we get:
      • K e = [PRODUCTS] / [REACTANTS] at equilibrium for the given reaction then is:
      • K e = ([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.
    • THE EQUILIBRIUM EXPRESSION a A + b B c C + d D K = [C] x [D] _________ [A] x [B] a b c d e
    • THE EQUILIBRIUM EXPRESSION
      • For a system at equilibrium, the value of the equilibrium constant (K e ) 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 H 2(g) + O 2(g)  2 H 2 O (g)
      • (the double arrow  means equilibrium)
      • K e = [H 2 O] 2 / ([H 2 ] 2 x [O 2 ])
      • How about H 2(g) + 1/2 O 2(g)   H 2 O (g ) ?
      • The reaction is the same but it is balanced with a different set of coefficients and therefore the equilibrium expression is different: K e = [H 2 O] / ([H 2 ] x [O 2 ]1/2 )
    • 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 H 2(g) + I 2(g)  2 HI (g) the equilibrium expression is
      • K e = [HI] 2 / ([H 2 ] x [I 2 ])
      • Let us assume the following equilibrium concentrations [H 2 ] = 0.094 M, [I 2 ] = 0.094 and [HI] = 0.012 M
      • K e = (0.012) 2 / ((0.094) x (0.094)) = 0.016
      • If the equation was balanced as ½ H 2(g) + ½ I 2(g)  HI (g) the equilibrium expression would be:
      • K e = [HI] / ([H 2 ] 1/2 x [I 2 ] 1/2 ) and
      • K e = (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!
    • HOW ARE K e VALUES RELATED?
      • In our previous example when the coefficients of the balanced equation were halved the K e value changed. But how?
      • H 2(g) + I 2(g)  2 HI (g) ,
      • K e = [HI] 2 / ([H 2 ] x [I 2 ]) = 0.016
      • ½ H 2(g) + ½ I 2(g)  HI (g) ,
      • K e = [HI] / ([H 2 ] 1/2 x [I 2 ] 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 H 2(g) + 2 I 2(g)  4 HI (g) , the constant would change to the square of the original value or 0.016 2 and K e would equal 0.064
      • When the balancing coefficients of an equation are changed by a factor, the K e value changes by that factor as a power of the original K e
    • HOW ARE K e VALUES RELATED?
      • How else may the original balanced equation be altered?
      • Often equations are reversed (interchanging products with reactants. For example:
      • H 2(g) + I 2(g)  2 HI (g) may be rewritten as:
      • 2 HI (g)  H 2(g) + I 2(g)
      • The equilibrium expression for the first equation is:
      • K e = [HI] 2 / ([H 2 ] x [I 2 ]) , for the second it is:
      • K e = ([H 2 ] x [I 2 ]) / [HI] 2 , the reciprocal of the first
      • Therefore when an equation is reversed, the K e becomes the reciprocal of the original K e value.
      • K e (reverse) = 1 / K (forward)
    • 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.
      • H 2(g) + I 2(g)  2 HI (g) (homogeneous – all gases)
      • H 2(g) + ½ O 2(g)  H 2 O (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.
    • 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 H 2(g) + O 2(g)  2 H 2 O (l)
      • K e = 1 / ([H 2 ] 2 x [O 2 ])
      • For a system such as:
      • CaCO 3(s)  CaO (s) + CO 2(g)
      • The equilibrium expression is:
      • K e = 1 x [CO 2 ] / 1 or simply K e = [CO 2 ]
      • notice that since both CaCO 3(s) and CaO (s) are solids they have been replaced by ones in the equilibrium expression.
    • THE MAGNITUDE OF K e 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). Ke = [ PRODUCTS ] [ REACTANTS ] __________ = a small value
    • THE MAGNITUDE OF K e 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). Ke = [ PRODUCTS ] [ REACTANTS ] __________ = a large value
    • THE MAGNITUDE OF K e AND EQUILIBRIUM CONCENTRATIONS When the equilibrium constant is about one, the Concentrations of the reactants and the products are about equal at equilibrium. Ke = [ PRODUCTS ] [ REACTANTS ] __________ = about one (1)
    • MEASURING K e 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 K e .
      • When the concentration values used in calculating K e are in terms of molar units the constant is referred to as K c (for concentration units – moles per liter ). When concentration of gases is measured in pressure units (atms) the constant is referred to as K p (for pressure units – atms).
      • The equilibrium constant in concentration terms (K c ) can be converted to an equivalent value in pressure units (K p ).
      • K p = K c (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.
      *
    • CONVERTING K P AND K C VALUES
      • Problem : For the reaction:
      • N 2(g) + 3 H 2(g)  2 NH 3(g) , K c = 0.105 at 472 0 C. Find K p at this temperature.
      • Solution:
      • K p = K c (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 NH 3(g) product gases) – ( one mole of N 2(g) and three moles of H 2(g) a total of four moles of reactant gases)
      • K p = 0.105 x (0.0821 x 745) -2 = 2.81 x 10 -5
      *
    • CALCULATING THE EQUILIBRIUM CONSTANT
      • Problem : If 0.20 moles of PCl 3(g) and 0.10 moles of Cl 2(g) are placed in a 1.0 liter container at 200 0 C the equilibrium concentration of PCl 5 is found to be 0.012 molar. What is K c for the reaction:
      • PCl 3(g) + Cl 2(g)  PCl 5(g)
      • Solution : Since one PCl 3 and one Cl 2 must be consumed to form one PCl 5 , (0.20 M, the starting molarity of PCl 3 – 0.012 M, the molarity of PCl 5 formed) = 0.188 M PCl 3 remains and
      • (0.10 M, the starting molarity of Cl 2 – 0.012 M the molarity of PCl 5 formed) = 0.088 M Cl 2 remaining ..
      • K c = [PCl 5 ] / ([PCl 3 ] x [Cl 2 ])
      • K c = (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!
    • USING THE EQUILIBRIUM CONSTANT
      • Problem: For the reaction: H 2(g) + I 2(g)  2 HI (g) , K c = 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: H 2(g) + I 2(g)  2 HI (g) and therefore,
      • K e = [HI] 2 / ([H 2 ] x [I 2 ])
      • The starting [H 2 ] and [I 2 ] 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 H 2 and 1 I 2 (half as many of each) are needed. The equilibrium [H 2 ] and [I 2 ] are both then (0.10 - .5X). This represents the starting concentrations of each minus the concentration used to form the HI.
    • USING THE EQUILIBRIUM CONSTANT (cont’d)
      • Substituting the equilibrium concentrations into the equilibrium expression we get:
      • K c = X 2 / (0.10 – 0.5 X)(0.10 – 0.5X) = 0.016 or
      • X 2 / (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 ,
      • [H 2 ] e = [I 2 ] 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 K c (0.016) considering the rounding errors!
    • USING THE QUADRATIC EQUATION WITH THE EQUILIBRIUM CONSTANT
      • Equilibrium calculations often require the use of the quadratic equation. Its function is to solve mathematical equations involving squared, first power and zero power terms in the same equation such as:
      • 4X 2 +0.0048X – 3.2 x 10 -4 = 0 this equation cannot be solved easily by inspection and requires the quadratic formula:
      • Using the form aX 2 + bX + c = 0 the formula is:
      • ( - b +  b 2 – 4ac )/ 2a
      • In the given equation: a = 4, b = 0.0048 and c = – 3.2 x 10 -4
      • (-0.0048 +  (0.0048) 2 – 4(4)( – 3.2 x 10 -4 )) / 2(4) = 0.0083
      • -0.0095
      • Note: every quadratic has two answers .
      *
    • USING THE QUADRATIC EQUATION WITH THE EQUILIBRIUM CONSTANT (cont’d)
      • Problem: Given the reaction
      • H 2(g) + I 2(g)  2 HI (g) , assume the original [H 2 ] = 0.200 M and [I 2 ] = 0.100 M. The K c value is 0.016 at 793 K. Find all equilibrium concentrations.
      • Solution: Let X = [H 2 ] consumed to reach equilibrium.
      • For every H 2 consumed so is one I 2 therefore X also = [I 2 ] consumed.
      • For every H 2 and I 2 consumed 2 HI are formed. Therefore [HI] e = 2X, and [H 2 ] e = (0.200 –X) and [I 2 ] e = (0.100 – X)
      • K e = [HI] 2 / ([H 2 ] x [I 2 ]) = (2X) 2 / (0.200 – X)(0.100 –X)
      • Kc = 4X 2 / (0.020 - 0.30X + X 2 ) = 0.016
      • 4X 2 = 0.016(0.020 - 0.30X + X 2 )
      • 4X 2 = 0.00032 - 0.0048X + 0.016 X 2
      *
    • USING THE QUADRATIC EQUATION WITH THE EQUILIBRIUM CONSTANT (cont’d)
      • 3.984X 2 + 0.0048X - 0.00032 = 0
      • a = +3.984, b = +0.0048, c = -0.00032
      • (- b +  b 2 – 4ac) / 2a
      • (-0.0048 +  (0.0048) 2 – 4(3.984)(-0.00032)) /2(3.984)
      • X = (-0.0048 + 0.0714) / 7.968 = 0.00833
      • -0.0095
      • NOTE : A negative X value means less than no H 2 was consumed which is of course impossible. Therefore the positive value must be correct!
      • [HI] e = 2 x 0.0083 = 0.017 M
      • [H 2 ] e = 0.200 – 0.0083 = 0.192 M
      • [I 2 ] e = 0.100 – 0.0083 = 0.0920 M
      • CHECK: (0.017) 2 / (0.192)(0.092) = 0.0164 ~ K c = 0.016
      *
    • REACTION QUOTIENTS AND THE DIRECTION OF EQUILIBRIUM
      • It is often important to determine whether a reacting system is at equilibrium, and if it is not, how will it change as it moves towards the equilibrium state.
      • In order to make this determination, a value called the reaction quotient. The reaction quotient is a mathematical setup just like the equilibrium expression except equilibrium concentrations are no necessarily used in its calculation.
      • Q = [products] / [reactants] but not necessarily at equilibrium!
      • The value of Q is compared the the equilibrium constant K e to decide if the reaction is already at equilibrium and in which direction the system will move to attain equilibrium.
      *
    • REACTION QUOTIENTS AND THE DIRECTION OF EQUILIBRIUM
      • If the value of Q > K e the reaction is not at equilibrium and will favor the reverse reaction to reach equilibrium.
      • If the value of Q < K e the reaction is not at equilibrium and will favor the forward reaction to reach equilibrium.
      • If the value of Q = K e the reaction is already at equilibrium and no net reaction will occur to reach equilibrium.
      *
    • REACTION QUOTIENTS AND THE DIRECTION OF EQUILIBRIUM
      • Problem: Given the following reaction:
      • 2 HF (g)  H 2(g) + F 2(g) , K e = 1 x 10 -13 at 1000 o C
      • [HF] = 0.50 M, [H 2 ] = 0.0010 M, [F 2 ] = 0.0040 M
      • Is the system at equilibrium? If not in which direction will it move to attain equilibrium?
      • Solution: Q = [HF] 2 / ([H 2 ] x [F 2 ])
      • Q = (0.50) 2 / (0.0010 x 0.0040) = 1.6 x 10 -5
      • Since Q does not equal K e the system is not at equilibrium
      • Q > K e therefore the system will move in reverse until equilibrium is reached. HF will be formed and H 2 and F 2 will be consumed!
      *
    • LeChatelier’s Principle and Chemical Equilibrium
      • LeChatelier’s Principle how equilibrium systems change when equilibrium conditions are changed.
      • The changes imposed on an equilibrium are called stresses . The response to these changes (stresses) are called shifts.
      • Stresses consist of changes in concentrations, pressures and temperatures . Shifts that occur are characterized as “shift towards products” (shift right) which means increased rate of forward reaction occurs or “shift towards reactants” (shift left) which means increased rate of reverse reaction occurs.
      • LeChatelier’s Principle defines the relationship between applied stresses and equilibrium shifts which result.
    • LeChatelier’s Principle and Chemical Equilibrium
      • In the reaction of A + B  C + D if the concentration of a reactant (A or B) increases, the collision rate between reactant molecules increases and the forward rate therefore increases. As the reaction continues and reactants are consumed the forward rate slows. At the same time more product is formed and the reverse reaction rate increases. At some point the rates forward and reverse again become equal and a new equilibrium is reached. At this new equilibrium however the concentrations of products are larger and the concentrations reactants are lower than in the original reaction. This is an equilibrium shift toward products.
    • LeChatelier’s Principle and Chemical Equilibrium A + B C + D A stress is applied, the concentration of reactant A is increased A C + D Equilibrium shifts forward and concentrations of products (C and D) increases. Concentration of reactant B is reduced because it must be consumed to allow the shift forward . The concentration of A is of course larger than the original because it was added although some of it is consumed in the forward shift. The equilibrium constant value remains unchanged despite the shift!
    • LeChatelier’s Principle and Chemical Equilibrium
      • In the reaction of A + B  C + D if the concentration of a product (C or D) increases, the collision rate between product molecules increases and the reverse rate therefore increases. As the reaction continues and products are consumed the reverse rate slows. At the same time more reactant is formed and the forward reaction rate increases. At some point the rates forward and reverse again become equal and a new equilibrium is reached. At this new equilibrium however the concentrations of reactants are larger and the concentrations products are lower than in the original reaction. This is an equilibrium shift toward reactants.
    • LeChatelier’s Principle and Chemical Equilibrium A + B C + D A stress is applied, the concentration of product C is increased Equilibrium shifts reverse and concentrations of reactants (A and B) increases. Concentration of reactant D is reduced because it must be consumed to allow the shift reverse . The concentration of C is of course larger than the original because it was added although some of it is consumed in the reverse shift. The equilibrium constant value remains unchanged despite the shift! C A + B
    • A + B C + D B LeChatelier’s Principle and Chemical Equilibrium A stress is applied, the concentration of product C is decreased C The reaction rate in the forward direction remains unchanged because reactant concentrations are not changed. The rate of the reverse reaction slows due to the reduced concentration of C and the net reaction shifts forward making more D and also increasing C while consuming reactants A and B. The value of the equilibrium constant remains unchanged . A D C
    • LeChatelier’s Principle and Chemical Equilibrium
      • When changing temperature is the applied stress, the exothermic or endothermic nature of the reaction must be considered.
      • When heat is added to endothermic reactions they shift forward. When heat is removed from endothermic reactions they shift in reverse.
      • When heat is added to an exothermic reactions they shift reverse. When heat is removed from exothermic reactions they shift forward.
      • Endothermic reactions are identified with a positive  H or heat written into the equation as a reactant. Exothermic reactions are identified with a negative  H or heat written into the equation as a product.
    • LeChatelier’s Principle and Chemical Equilibrium A + B C + D + heat Exothermic reactions shift reverse when heat is added Concentrations of A and B increase, C and D decreases heat A + B A + B C + D + heat heat A + B C + D Endothermic reactions shift forward when heat is removed Concentrations of C and D increases, A and B decreases.
    • LeChatelier’s Principle and Chemical Equilibrium
      • In general, Le Chatelier’s Principle tells us that equilibrium systems when disturbed by an applied stress, shift away from increased concentrations and towards reduced concentrations.
      • Similarly, systems shift away from added heat (increased temperatures) and towards removed heat (lower temperatures).
      • It is important to note that altering concentratio ns shift equilibriums but DO NOT CHANGE the value of the equilibrium constant . Changing temperatures cause equilibrium shifts but DO CHANGE the value of the equilibrium constant.
    • LeChatelier’s Principle and Chemical Equilibrium
      • In addition to concentrations and temperatures often pressures are changed in an equilibrium system.
      • LeChatelier’s Principle can be applied to these system also. When gaseous systems are compressed the equilibrium shifts towards the side of the reaction containing the fewer number of gas phase molecules.
      • When gaseous systems are expanded, the equilibrium shifts towards the side of the reaction containing the greater number of gas phase molecules
      • For example: N 2(g) + 3 H 2(g)  2 NH 3(g)
      • compressing this system would cause a shift to the right (products) which contains only two gas phase molecules (2 ammonia NH 3 ).
      • Expanding this system would cause a shift to the left (reactants) which contains four gas phase molecules (1 nitrogen and 3 hydrogen)
    • LeChatelier’s Principle and Chemical Equilibrium
      • Problem: Given the following reaction:
      • 2 H 2(g) + O 2(g)  2 H 2 O (l) + heat
      • How can the amount of water formed be increased?
      • Solution:
      • Increasing the concentration of hydrogen will cause an equilibrium shift forward.(more water formed)
      • Increasing the concentration of oxygen will cause an equilibrium shift forward. (more water formed)
      • The reaction is exothermic . Removing heat (lowering the temperature) will cause a shift forward.
      • Compressing the system will cause a shift forward (two gas phase molecules convert to zero gas phase molecules)
      • Removing liquid water from the system will not cause a equilibrium shift. Remember that liquids and solids are not involved in equilibrium systems!
    • SUMMARY OF KEY IDEAS
      • (1) At equilibrium
      • Rate forward Reaction = Rate reverse Reaction
      • (2) For the reaction: a A + b B  c C + d D
      • K e = ([C] c x [D] d) / ([A] a x [B] b )
      • (3) The numerical value of the equilibrium constant depends on the coefficients used to balance the equation!
      • (4) When the balancing coefficients of an equation are changed by a factor, the K e value changes by that factor as a power of the original K e
      • (5) Therefore when an equation is reversed, the K e becomes the reciprocal of the original K e value.
      • (6) Since the equilibrium expression involves only components which change concentration, solids and liquids are not included in its format.
    • SUMMARY OF KEY IDEAS
      • (7) Large equilibrium constants mean products are favored at equilibrium. Small values mean reactants are favored. Constants of about one mean products and reactants are present in approximately equal concentrations at equilibrium
      • (8) K p = K c (R x T)  n (converting K p to K c )
      • (9) The quadratic formula is sometimes required for problem solution: aX 2 + bX + c = 0
      • ( - b +  b 2 – 4ac )/ 2a
      • (10) Q = [products] / [reactants] but not necessarily at equilibrium!
      • (11) Q> K e (reactants favored), Q< K e (products favored), Q = K e (the system is at equilibrium)
    • SUMMARY OF KEY IDEAS
      • (12) LeChatelier’s Principle
      • Concentration of reactants increases, shift towards the products
      • Concentration of products increases, shift towards the reactants
      • Exothermic reactions, add heat, shift toward reactants
      • Endothermic reactions, add heat, shift toward products
      • Compress a gaseous system, shift toward fewer gas molecules
      • Expand a gaseous system, shift toward more gas molecules
    • THE END