Chemical Reactions: 4-Thermodynamics - S11
Chemical Reactions: 4-Thermodynamics - S11
Chemical Reactions: 4-Thermodynamics - S11
DOC
CHEMICAL REACTIONS 1. The Rate Law (SJ p. 27, Chap. 2.4) - The principle of mass action (The law of mass action) k1 k2
aA + bB
cC
+ dD
where k1 = rate constant for the forward reaction k2 = rate constant for the reverse reaction Rate of forward rxn = k1 {A}a {B} b Rate of reverse rxn = k2 {C}c{D}d 3. Effect of Temperature on the Reaction Rate (SJ p.40-45, chap.2.6, 2.7) a. Experiments have shown that chemical reaction rates increase with increasing temperature. Rate of forward rxn = k1 {A}a {B} b Rate of reverse rxn = k2 {C}c{D}d
as T as T
b. In many instances, the effect of temperature on reaction rate is related to its effect on the reaction rate constant. k1 = f (T) k2 = f (T) { }= f (T)
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Arrhenius Law (SJ p. 40, Effect of Temperature on Reaction Rate) Arrhenius formulated the empirical rate law: - Ea k = A exp (------) RT (kcal/mol) ---------------------(kcal/mol K) (K)
(1)
where k = the rate constant A = the pre-exponential factor, or frequency factor (A is usually treated as a constant that is independent of temperature for a particular reaction.) Ea = the activation energy, kcal/mol (Ea is also treated as a constant for a particular reaction) R = universal gas constant for ideal gas or the ideal gas constant = 1.9872 cal/mol K (= 1.99 cal/mol K) = 0.08205 Latm/mol K T = the absolute temperature in K Note: 1 calorie = 4.184 Joul Taking logarithm of (1), it can be linearlized as: Ea ln k = ln A -----RT or Ea 1 ln k = ---- ---- + ln A R T
(2)
Arrhenius Plot
ln A
A plot of ln k vs. 1/T should give a straight line with: a slope = Ea/R an intercept = ln A
ln k
slope = - Ea/R
1/T
4-Thermodynamics_S11.DOC
At T1,
Ea ln k1 = -----RT1 Ea ln k2 = -----RT2
ln A
(3)
At T2,
ln A
(4)
Subtract (3) from (4): Ea 1 1 ln k2 ln k1 = ------ (----- -----) R T2 T1 k2 ln -----k1 Ea or ---------------- = ------1 1 R ---- ---T2 T1
(5)
k 1 Ea T2 T1 Ea 1 ln 2 = = k1 R T1 T2 R T1T2
kT = k20C (T 20C)
where kT = the rate constant at a temperature T (C) Example: For deoxygenation rate constant, kT = k20C 1.048 (T 20C) For reaeration rate constant, kT = k20C 1.024 (T 20C) Q10 effect In the field of biology, the term Q10 is frequently used. k(T + 10C) Q10 = -------------- 2 kT where kT = C
4-Thermodynamics_S11.DOC
Arrhenius Formula
Ea RT
k = Ae
(1)
k 1 = Ae
Ea RT1
at T1 at T2
(2) (3)
k 2 = Ae
Ea RT2
(4) (5)
ln k 2 ln k1 =
Since
(6)
Eq (6) is rewritten as
k Ea 1 1 Ea T2 T1 ln 2 = = k1 R T1 T2 R T1T2
(7)
This is the Arrhenius formula for temperature correction on reaction rate constant.
4-Thermodynamics_S11.DOC
Effect of Temperature on the Energy Level Distribution of Reacting Species The Eyring's Transition State Theory We assume that the reaction proceeds through a high-energy, unstable intermediate known as a transition complex or an activated complex, with a release of energy, H, the heat of reaction.
Ea Unstable intermediate Activated complex or Transition complex Ea H
A + B
Products
Extent of reaction
a) The intermediate formation of the activated complex requires the input of an average amount of energy Ea, the activation energy of the forward reaction. b) The energy Ea as well as H is released when the activated complex decomposes into the products of the reaction. c. For the reverse of this reaction to proceed, an energy input of (H + Ea) is required to form the activated complex. (H + Ea) is the activation energy for the reverse reaction.
4-Thermodynamics_S11.DOC 3. The Maxwell-Boltzmann theory a. The effect of temperature on the distribution of energy levels for a given species can be described by The Maxwell-Boltzmann theory E N = No D exp (- -----) RT where E = a stipulated energy level No = total number of molecules N = number of molecules with energy equal to or greater than E R = gas constant T = absolute temperature, K D = constant Fig. 2.5 Reactant energy distribution
N Number of molecules with stated energy
b. For given T, N will decrease as as E increased. As the stipulated energy level is raised, a smaller and smaller fraction of the total molecules are increased in N. c. For a given energy level, increasing T should exponentially increase the population of molecules present at or above this energy level.
E Energy E
d. Only molecules with an energy level equal to or greater than E are capable of forming the high energy level transition complex.
CHEMICAL EQUILIBRIUM k1 aA + bB k2 vf = k1 {A}a {B}b vr = k2 {C}c{D}d Rate of forward rxn Rate of reverse rxn cC + dD
1. Equilibrium (SJ p.58-60, chap. 3.1-3.2) At equilibrium, vf = vr = k1 {A}a {B}b = k2 {C}c{D}d k1 {C}c{D}d Keq = ----- = ------------{A}a {B}b k2 2. The thermodynamic Basis of Chemical Equilibrium (SJ, p.60, chap. 3.3) - used to answer the question "Will this reaction go?" 1. The Position of Chemical Equilibria a. By knowing the position of chemical equilibria, we can determine whether it is possible for certain reactions between reactants at given concentrations to proceed. b. We can provide answers to questions "Will this reaction go?" such as: Will calcium carbonate tend to precipitate or dissolve in this water? Can I possibly oxidize sulfide with nitrate? c. There are two general ways to answer questions like these: 1) An experiment 2) Calculate the answer using previously determined equilobrium data.
4-Thermodynamics_S11.DOC
Thermodynamics Free Energy, G a. Thermodynamic expression for free energy: G Energy content = H Heat content TS State of randomness
where G = Gibbs free energy, kcal T = absolute temperature, K H = enthalpy, kcal S = entropy, kcal/K Free Energy, G - part of the total energy that is available to perform "useful work", other than pressurevolume work. Enthalpy, H - total energy content of an element or compound. Entropy, S - a sort of internal manifestation of energy. e.g., degree of order or organization in a system - Highly structured materials (e.g., well-formed crystals) have low entropy - Randomly arranged systems (e.g., a gas) have high entropy. TS - part of the total energy which is not available for useful work - degree of organization of the system - state of randomness
Ludwig von Boltzmann (1906) S = K ln W where K = Boltzmann constant W = probability state should occur
- Greater the probability, greater the entropy "S = K ln W" is written in his grave stone. He suicide, it was proven one year after his death.
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Total Free Energy, GT (SJ, p. 61) Total free energy is the sum of the free energies of each of the reaction components. For the reaction, aA + bB cC + dD
__ __ __ __ GT = nA GA + nB GB + nC GC + nD GD where nA, nB, nC, and nD = number of moles of A, B, C, and D, respectively __ __ __ __ GA, GB, GC, and GD = free energy per mole of A, B, C, and D, respectively Criterion for Equilibrium Consider the above reversible reaction. If we were to add A and B to a reactor vessel and calculate the total free energy of the system as a function of extent of reaction as the reaction proceeded, we would find something like this:
Only reactants A & B present Total free energy, GT Only products C & D present
GT decrease
Extent of reaction
Fig. 3.2 Variation of Gibbs free energy for the chemical reaction: aA + bB cC + dD
- Only reactants are present at the far left of the diagram and only products at the far right side. - Conversely, if we were to add only C and D to the reaction vessel, calculation of the total free energy of the system as the reaction proceeded to form A and B would produce a curve of the form shown by the dashed line on the right-hand side. - In each case, the reaction proceeds spontaneously (or without any external help), as long as the value of GT decreases.
For closed systems: At constant pressure and constant temperature, the criterion for equilibrium is that the total free energy of the system is minimum: GT = minimum.
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It may be stated that: a) The equilibrium condition of reaction is the point at which GT is a minimum. b) The reaction in the direction that decreases GT is spontaneous. c) The reaction in the direction that increases GT is not spontaneous or will not occur in a closed system.
Chang in Total Free Energy, GT - For any reactions proceed an incremental amount, the change in GT is proportional to G.
G = i Gi i Gi i products i reactants
where i = stoichiometric coefficient (e.g., a, b, c, and d) Gi = free energy per mol (kcal/mol) We may state that: If G < 0, GT decreases, the reaction may proceed spontaneously as written. If G > 0, GT increases, the reaction cannot proceed spontaneously as written. If G = 0, GT is minimum, the reaction is at equilibrium and the reaction will not proceed spontaneously in either direction.
The Thermodynamics Basis of Chemical Equilibrium (SJ, 60) General eqn for the reaction: G = G + RT ln Q where Q = reaction quotient aA + bB cC + dD
G = i Gi i Gi i products i reactants
where = Gi = free energy per mol of species i at standard state conditions (i.e., 20 C, 1 atm)
4-Thermodynamics_S11.DOC
Note: By convention, every element is assigned free energy of zero at standard state; i.e., H2(g), O2(g), C graphite(s) are all Gi = 0 kcal/mol.
The first step in determining G is to determine, Gi , the free energy per mole of reactant and product at the standard state condition.
If chemical equilibrium applies, then G = 0 and Q = Keq G = - RT ln Keq G Note: ln Keq = -------RT Substituting (2) into (1) yields G = - RT ln Keq + RT ln Q Q G = RT ln -----Keq This gives important relationship between thermodynamics and equilibrium. If Q/Keq < 1, then G < 0 If Q/Keq > 1, then G > 0 If Q/Keq = 1, then G = 0 (G = -): the reaction is spontaneous as written. (G = +): the reaction cannot proceed as written. : the reaction is at equilibrium. (1) (2)
Example: Calculate the standard free energy change, G, for the reaction in which calcite CaCO3(s) dissolves in acidic water to form Calcium Ca2+ and bicarbonate HCO3- ions: CaCO3(s) + H+
Ca2+
HCO3
4-Thermodynamics_S11.DOC
+ H+ 0 _
Ca2+
+ HCO3 -140.31
-269.78 _
-132.18
i G ) i G ) G = ( i products ( i reactants i i
= [(-132.18 + (-140.31)] [(- 269.78) + 0] = - 2.71 kcal/mol
{C} {D} In order to determine if the reaction will proceed, we need to calculate Q = a b { A} {B}
since G = G + RT ln Q From Jackson and Patterson (1982), at the piezometer M2, Ca 2+ = 12.8 mg/L = 0.6384 meq/L = 0.3192 mM = 0.3192 x10-3 M HCO3 - = 0.4 meq/L = 0.40 mM = 0.40 x10-3 M pH =6, H+ =1x10-6 M Note that activity of solids =1 and assume, = 1 and T = 15C (273+15 =288K) {Ca2+}{HCO3 } (0.3192 x10-3 )(0.40 x10-3) Q = ----------------------= --------------------------------- = 0.12768 {CaCO3(s)}{H+ } (1 )(1x10-6) G = G + RT ln Q = - 2.71 kcal/mol + (1.9872 x 10-3 kcal/mol K) (288 K) ln (0.12768) = - 3.888 kcal/mol G < 0, the reaction proceeds spontaneous as written.
Note that, in Jackson and Patterson (1982), the concentration of Ca2+ and HCO3- increased gradually in groundwater.
Example 3-1 (SJ 68) H2O(l) dissociates to H+ and OH- at 20 C: H2O(l) H+ + OH-
1) Determine the equilibrium constant, Keq, for the reaction. 2) Is this reaction proceeding as written when [H+] = 1.000 x10-6 and [OH-] = 5.000 x10-8 ? Ignore the ionic strength effect (assume H+ = OH- = 1)
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H2O(l) -56.690
H+ 0
OH-37.595
G = ( i Gi )products ( i Gi )reactants
i i
= [0 + (-37.595)] (-56.690) = 19.095 kcal/mol (Note 1 mol of H2O(l) reacted) At equilibrium, G = - RT ln Keq At 25 C, 19.095 kcal/mol = - (1.9872 x 10-3 kcal/mol K) [(273+25) K] ln Keq = - (0.593 kcal/mol) ln Keq ln Keq = - (19.095 / 0.593) = -32.2 Keq = e-32.2 = 1 x 10-14 = Kw Note: { H+}{ OH-} Keq = ---------------- = { H+}{ OH-} = Kw = 1x10-14 { H2O } {H2O} = 1 2) G = G + RT ln Q where Q = reaction quotient {H+}{OH-} Q = ---------------- = {H+}{OH-} = {1x10-6}{5x10-8} = 5x10-14 {H2O} Note {H2O} = 1 G = G + RT ln Q G = 19.095 kcal/mol + (1.9872 x 10-3 kcal/mol K)(298 K) ln (5x10-14 ) = +0.93 kcal/mol at 25 C
4-Thermodynamics_S11.DOC
Because G > 0, the reaction is not spontaneous as written, and can proceed spontaneously only in the opposite direction; that is, H+ and OH- are combining to form H2O molecules. Note: Q G = RT ln -----Keq Q 5x10-14 ---- = ---------- = 5 > 1 Keq 1x10-14
Enthalpy, H G = H - TS H = the enthalpy change of a chemical reaction = the amount of heat that is released or taken up during the course of the reaction. If H < 0, heat is evolved If H > 0, heat is taken up (exothermic) (endothermic)
Exothermic reaction reaction that give off energy (have negative enthalpy) Endothermic reaction reaction that absorb energy (have positive enthalpy) and require heat or electrical energy to begin.
i H ) i H ) H = ( products ( reactants i i i i
where i = stoichiometric coefficient H i = enthalpy of species i (kcal/mol) at 25C and 1 atm. By convention, in aqueous solution, 1 M of H+ in an ideal solution ( H+ = 1) is assigned H i = 0 kcal/mol. __ Hf = the enthalpy of formation
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4-Thermodynamics_S11.DOC
Example (Lindeburg, p. 35-5) What is the enthalpy of reaction at 25 C for the combustion of ethane (C2H6 or CH3CH3) ? 2 C2H6 + Hf Hf Hf Hf of of of of 7 O2 C2H6 O2 CO2 H2O > 4 CO2 + 6 H2O
(A) -680 kcal/mol (B) -340 kcal/mol (C) 130 kcal/mol (D) 340 kcal/mol
(Solution) 2 C2H6 _ H (kcal/mol) -20.24 + 7 O2 0.00 > 4 CO2 -94.05 + 6 H2O -57.80
The enthalpy of reaction is H = {Hf, products} - {H f, reactants} = [(4 mol) (- 94.05 kcal/mol) + (6 mol)(-57.80 kcal/mol) ] - [ (2 mol)(-20.24 kcal/mol) + (7 mol)(0.00 kcal/mol)] = - 682.5 kcal (Exothermic) for 2 mol of ethane (C2H6) For 1 mol of ethane, the enthalpy of reaction is ~340 kcal/mol. Answer is B.
Organic Chemistry - Alkane General Formula: RH No. of Molecular Carbon Formula 1 CH4 2 CH3 CH3 3 CH3 CH2 CH3 4 CH3 (CH2)2 CH3 5 CH3 (CH2)3 CH3 6 CH3 (CH2)4 CH3 7 CH3 (CH2)5 CH3 8 CH3 (CH2)6 CH3 9 CH3 (CH2)7 CH3 10 CH3 (CH2)8 CH3 Functional Group: C-H and C-C bonds Molecular Name Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane
4-Thermodynamics_S11.DOC
Example: Calculate enthalpy change (heat) of the reaction of lactic acid to pyruvic acid: lactic acid pyruvic acid __ H (kcal/mol) CH3COCOOH + 5/2 O2(g) 3CO2(g) + 2H2O -279.0 Pyruvic acid
-326.0
(Solution) Since the reaction occurs from lactic acid to pyruvic acid, H (kcal/mol) 3CO2(g) + 2H2O CH3COCOOH + 5/2 O2(g) +279.0 CH3CHOHCOOH + 3 O2(g) 3CO2(g) + 3H2O -326.0 _____________________________________________________________ CH3CHOHCOOH + O2(g) CH3COCOOH + H2O -47.0 kcal/mol H = - 47.0 kcal/mol (Exothermic)
Organic Chemistry Organic acid , Carboxylic Acid Functional group -C=O \ OH General formula R-C=O \ OH Example compound CH3-C=O \ OH
CH3-C=O + H+ \ O-
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Alkyl radicals, R No. of Radical Carbon Formula 1 CH3 2 CH3 CH2 3 CH3 CH2 CH2 4 CH3 (CH2)2 CH2 5 CH3 (CH2)3 CH2 6 CH3 (CH2)4 CH2 7 CH3 (CH2)5 CH2 8 CH3 (CH2)6 CH2 9 CH3 (CH2)7 CH2 10 CH3 (CH2)8 CH2 Radical Name Methyl Ethyl Propyl Butyl Pentyl (Amyl) Hexyl Heptyl Octyl Nonyl Decyl
IUPAC* Hexanoic acid Heptanoic acid Octanoic acid Nonanoic acid Decanoic acid
4-Thermodynamics_S11.DOC
Temperature Effect on the Equilibrium Constant - Temperature dependence of the equilibrium constant G = H - TS G = - RT ln Keq From (2), ln K eq =
G RT Taking derivative of both sides with respect to T
(1) (2)
(2)
d ln K eq dT
G 1 2 R T
Note:
d dT 1 1 d T 1 = 1 T 2 = 2 = T T dT
( )
( )
d ln K eq dT
G RT 2
K2
K1
d ln K eq =
T2
T1
H dT RT 2
1 1 1 2 = T = T2 T1 T
T
T2
T1
K H 1 ln 2 = K1 R T
T1
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4-Thermodynamics_S11.DOC
ln
K1 H 1 1 = K2 R T2 T1
ln
ln
K1 H 1 1 = K2 R T2 T1
or
K2 H T1 T2 = K1 R T1T2
Example 6-2 (SJ 251) Find the equilibrium constant (solubility product, Ksp) for FePO4(s) at 50C given the following information at 25 C for the reaction:
FePO4(s)
Fe 3+
+ PO4 3-
For the above reactions at 25C, H = -18.7 kcal/mol and G = 24.4 kcal/mol. Note: The reaction shows that FePO4(s) dissolve in water to form Fe3+ and PO4 3- . The equilibrium constant is called the solubility product, Ksp. (Solution)
K sp ( 25C ) = e
ln
=e
= 1.27x10-18 = K1
K1 H 1 1 = K2 R T2 T1 1.27 x1018 1 18.7 kcal / mol 1 = 3 K2 1.987 x10 kcal / mol K 323K 298 K = 2.444
ln
= 1.10 x1019
At 25C, K1 = 1.27x10-18 At 50C, K2 = 1.10x10-19 Ksp decreases as temperature increases; thus solubility of FePO4(s) decreases as temperature increases.