Reaction rate ChE 400 - Reactive Process Engineering reaction rate describes change in concentration of reactants and products with time -> r = dc j /dt r is proportional to the reactant concentrations taken to some power the exponent of the power law is the order of the reaction in this reactant the order of the reaction is generally NOT equal to the stoichiometric coefficient of this reactant! The progress of a reaction can be denoted by the reaction extent or the (fractional) conversion r = k S j= 1 m j C j L3-1
Reaction Rate Coefficient k ChE 400 - Reactive Process Engineering Rate law: While k is for historic reasons often called a rate constant, it is in effect anything but constant - hence it is nowadays better called rate coefficient. It follows the famous Arrhenius law (or also Polanyi-Wigner relation): k(t) k(t) = k 0 is the pre-exponential factor (or sometimes also frequency factor), while E a is the activation energy of the reaction (and R and T the universal gas constant and the reaction temperature, respectively). energy E a reactant r = k S j= 1 product ΔH R m j C j power-law expression m j is the reaction order generally, m j ν j E a is a is We We will will also also assume assume that that kk o is o is L3-2 reaction coordinate
Exercise: ChE 400 - Reactive Process Engineering L3-3 As mentioned, k is anything but constant. While it does not depend on the reactant concentrations, it does depend very strongly (in fact: exponentially) on temperature. Assuming an activation energy of about 83 kj/mol, how much does the reaction rate change when the reaction temperature is increased from room temperature by 10 K (i.e. about 300 K to 310 K). { } k exp E kt = RT { } k exp E RT kt ( 2) ( ) 1 0 0 2 1 = With the given values: k(t 1 )/k(t 2 ) We get an increase in reaction rate by simply stepping up the temperature by 10 K. => How about 300K -> 350K?
Reversible Reactions ChE 400 - Reactive Process Engineering L3-4 In typical chemical processes, we are rarely dealing with irreversible reactions, i.e. a reaction which can only proceed in one direction. For a reversible reaction, the reaction rate has to be written as a difference between forward reaction r f r = r r = f b and reverse (or back) reaction r b : When a reversible reaction approaches its thermodynamic equilibrium, the rates of the forward and the backward reactions become equal and hence the net reaction rate becomes 0: r eq = r f r b = k f k = b Equilibrium does NOT mean that no reactions are taking place. Rather, the rates of the forward and back reactions are exactly the same!
Chemical Equilibrium ChE 400 - Reactive Process Engineering L3-5 As you (should have) learned in thermo, a chemical system assumes a state of minimum free enthalpy at equilibrium. This can be shown to lead to: S S S G Δ G = ν G = ν = ν μ = R j j j j j j= 1 j= 1 N j j= 1 TP, Substituting the definition of the chemical potential μ: we can write at equilibrium (how did we get here??): 0 0 ν S j μ j ΔG R ν j exp = exp = aj RT RT j= 1 This defines the equilibrium constant K eq. 0 μ = μ + K RT ln a 0 j j j To simplify our lives, we will restrict ourselves to ideal liquid solutions as well as ideal gases throughout this course and hence can replace a j by C j and P j, as needed. K eq ΔG = exp = = S S R RT j= 1 j= 1 eq
Excursion/Repetition: Thermo ChE 400 - Reactive Process Engineering Heat of reaction: (enthalpy of reaction) Exothermic reaction: ΔH R < 0 Endothermic reaction: ΔH R > 0 Δ H = ν H ( T) R j j j H j (T) = H j,f + C p,j (T) dt H j,f + C p,j ΔT Entropy of reaction: S j (T) = S j,f + C p,j (T)/T dt S j,f + C p,j ln(t 2 /T 1 ) Free (Gibb s) Enthalpy of reaction: Δ S = ν S ( T) R j j j ΔG R = ΔH R T ΔS R L3-6
Quick Example ChE 400 - Reactive Process Engineering L3-7 Example: N 2 + O 2 -> 2 NO (NO formation in air) Need thermodynamic data!? An excellent source is the NIST webbook (http://webbook.nist.gov). What is the heat of reaction? Δ H = ν H ( T ) = H H + 2H f f f f R j j N 2 O 2 NO j = What is the entropy of reaction? Δ S = ν S ( T ) = S S + 2S R j j N2 O2 NO j = What is the Gibbs Free Enthalpy of reaction at 300K? ΔG R = ΔH R T ΔS R =
T-dependence of K eq ChE 400 - Reactive Process Engineering L3-8 Another important relation that can be deduced from the above is the van t Hoff-equation, which describes the temperature dependence of the equilibrium constant: d K H dt RT 0 ln eq Δ R = 2 What follows for endothermic vs exothermic reactions from this equation?? It is often a good approximation to neglect the temperature-dependence of ΔH R in this equation. A typical plot of ln K eq vs 1/T hence yields a straight line. ln K 1 / T What can you deduce from the slope of the line? (Check LDS fig. 2-11, 2 p. 59)
ChE 400 - Reactive Process Engineering L3-9 Equilibrium Constants: Typical Values Equilibrium constants can take on VERY large values as well as VERY small ones. However, K eq is always > 0!
Equilibrium: Example 10ChE 400 - Reactive Process Engineering L3-10 Let s look at an example: decomposition of dinitrogen tetroxide to nitrogen dioxide: N 2 O 4 <=> 2 NO 2 We can deduce: Increasing shifts reaction equil.: T p Le Le Chatelier s Principle: : N 2 O 4 (colorless) NO 2 (brown) A system system reacts reacts to to a change change in in conditions in in a way way to to counteract this this change. Assuming ideal gas behavior, we can write for K eq : S 0 ( / ) S ν j eq j j j= 1 j= 1 K a P P ν j = = =
Le Chatelier s Principle 11ChE 400 - Reactive Process Engineering L3-11 Effect of some disturbances on an equilibrium system: Disturbance Reactant concentrations increase decrease System Pressure increase decrease Temperature increase decrease Net direction of reaction Effect on K eq
12ChE 400 - Reactive Process Engineering L3-12 Reaching Equilibrium A look at the molecular level: No change in time any more: equilibrium reached. colorless
The Reaction Quotient Q 13ChE 400 - Reactive Process Engineering L3-13 In a completely analogous way to the equilibrium constant K eq, we can define a reaction quotient Q: Q S = j= 1 j C ν j This coefficient, however, is formed with the currently present reactant and product concentrations. Comparison of Q with K eq tells us, which direction a reaction is going to proceed from the current state: Q < K eq -> reaction is proceeding from left to right (forward) Q > K eq -> reaction is proceeding from right to left (backward)
14ChE 400 - Reactive Process Engineering L3-14 Reaction Quotient vs Equilibrium Constant The reaction quotient changes with time while the equilibrium constant is time-independent. At equilibrium, the two values become identical.
Multiple Reactions: Q & K eq 15ChE 400 - Reactive Process Engineering L3-15 Problem: Remember NO x formation? Oxygen gas combines with nitrogen gas in the internal combustion engine to produce nitric oxide, which when out in the atmosphere combines with additional oxygen to form nitrogen dioxide. (1) N 2 (g) + O 2 (g) 2 NO (g) K c1 = 4.3 x 10-25 (2) 2 NO (g) + O 2 (g) 2 NO 2 (g) K c2 = 6.4 x 10 9 (a) Show that the overall Q c for this reaction sequence is the same as the product of the Q c s for the individual reactions. (b) Calculate K c for the overall reaction. Procedure: We first write the overall reaction by adding the two reactions together and write the Q c. We then multiply the individual K c s for the total K. (1) N 2 (g) + O 2 (g) 2 NO (g) (2) 2 NO (g) + O 2 (g) 2 NO 2 (g) overall: N 2 (g) + 2 O 2 (g) 2 NO 2 (g)
Multiple Reactions, Cont d 16ChE 400 - Reactive Process Engineering L3-16 N 2 (g) + 2 O 2 (g) For the individual steps: 2 NO 2 (g) Q (overall) = (1) N 2 (g) + O 2 (g) 2 NO (g) Q c1 = [NO 2 ] 2 [N 2 ][O 2 ] 2 (2) 2 NO (g) + O 2 (g) 2 NO 2 (g) Q c2 = [NO] Q c1 Q c2 = 2 [NO 2 ] 2 = [N 2 ] [O 2 ] [NO] 2 [O 2 ] (b) hence we can directly calculate: K = K 1 K 2 = [NO 2 ] 2 [N 2 ][O 2 ] 2 K eq for a sum reaction equals the product of K eq of the reaction steps!
NO 2 /N 2 O 4 equil. at 100 o C 17ChE 400 - Reactive Process Engineering L3-17 Back to the decomposition of dinitrogen tetroxide to nitrogen dioxide: N 2 O 4 <=> 2 NO 2 Initial conc. Equil. conc. Equil. Const. [N 2 O 4 ] [NO 2 ] [N 2 O 4 ] [NO 2 ] 0.1000 0.0000 0.0491 0.1018 0.0000 0.1000 0.0185 0.0627 0.0500 0.0500 0.0332 0.0837 0.0750 0.0250 0.0411 0.0930 [NO 2 ] 2 0.211 0.212 0.211 0.210 [N 2 O 4 ] It does not matter, from which side we approach the equilibrium!
Calculating equilibrium compositions 18ChE 400 - Reactive Process Engineering L3-18 From the definition of the equilibrium constant 0 S ΔG Ri νij Keq, i = exp = aj i= 1... R RT j= 1 We can directly calculate the equilibrium composition for a given reaction system. All we need is thermodynamic data for all reactants (-> NIST webbook, http://webbook.nist.gov). For a multiple reaction system (as written above) this will yield a system of R polynomials that need to be solved analytically. While this is not overly difficult, it is in practice done by computer programs, such as CHEMEQ, STANJAN (and many others). It is also often contained in large chemical process simulation software, such as ASPEN. (A publicly accessible webpage that is linked to a STANJAN routine can be found at http://navier.engr.colostate.edu/tools/equil.html)
Example: Ammonia Synthesis revisited 19ChE 400 - Reactive Process Engineering L3-19 Let s Let s try try to to put put this this all all together Remember ammonia synthesis?? N 2 + 3? H 2 <=> 2? NH 3 How is this reaction effected by pressure? K eq P y P P y y P 2 2 2 NH NH 1 3 3 = = 3 3 N H N H 2 2 2 2 How is it effected by temperature? H fo (NH 3 ) = -46 kj/mol What is H fo (N 2 ), H fo (H 2 )? How about being a little more quantitative? y 2 2 0 NH3 1 ΔG R Keq = 3 = exp yn y 2 H P 2 RT S N2 = 192 J/mol K ΔG = ΔH TΔS -> we need S S H2 = 130 J/mol K j! ΔS R = - 196 J/mol K S NH3 = 193 J/mol K
Example, cont d 20ChE 400 - Reactive Process Engineering L3-20 ΔH R = - 92 kj/mol ΔS R = - 196 J/mol K X A = N N N A0 A0 A ΔG R = ΔH R TΔS R -> ΔG R -> K eq (T) -> need to specify T So, now we know the equilibrium constant at a specified temperature. How do we get the conversion from here?? Which conversion? -> what is the limiting reactant? Let s assume stoichiometric feed of reactants: 3 N N2,0 = N H2,0 Let s pick X N2 (why does this not matter at all in this case??) From stoichiometry, we know: N NH3 = N NH3,0 + ν NH3 χ = 2 χ N H2 = N H2,0 3 χ = 3 N N2,0 3χ N N2 = N N2,0 - χ This can also be written in terms of conversion by simply filling in the definitions for reaction extent and conversion. Here, X N2 = χ / N N2,0
Example, cont d 21ChE 400 - Reactive Process Engineering L3-21 We now have number of moles as a function of conversion. That s still not what we need: we need mole fractions y j. y j = N j / N total N total = N N2 + N H2 + N NH3 = N N2,0 χ + 3 N N2,0 3χ + 2 χ = 4 N N2,0 2 χ = (4 2 X N2 ) N N2,0 N N2 = (1-X) N N2,0 y N2 = N N2 /N total = N H2 = (1-X) 3 N N2,0 y H2 = N H2 /N total = N NH3 = 2 X N N2,0 y NH3 = N NH3 /N total = K eq 2 2 y 0 NH 1 ΔG R exp 3 yn yh P RT 3 = = = 2 2 By simple iteration (or more fancy methods be my guest ), we can now directly calculate the conversion for given p,t.
22ChE 400 - Reactive Process Engineering L3-22 Example, cont d What would optimal reaction conditions for ammonia synthesis be??
23ChE 400 - Reactive Process Engineering L3-23 Example, last one So, why is ammonia synthesis industrially operated between 400 o C and 500 o C and at pressures between 150 and 300 atm rather than at room temperature and 1000 atm??