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MIT OpenCourseWare http://ocw.mit.edu 5.60 Thermodynamics & Kinetics Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.60 Spring 2008 Lecture #15 page 1 Chemical Equilibrium Ideal Gases Question: What is the composition of a reacting mixture of ideal gases? e.g. ½ N2(g, T, p) + 3/2 H2(g, T, p) = NH3(g, T, p) What are pN , pH , and pNH 2 2 3 at equilibrium? Let’s look at a more general case νA A(g, T, p) + νB B(g, T, p) = νC C(g, T, p) + νD D(g, T, p) The νi‘s are the stoichiometric coefficients. Let’s take a mixture of A, B, C, and D with partial pressures pA = XA p , pB = XA p , pC = XC p , and pD = XD p Is this mixture in equilibrium? We can answer by finding ∆G if we allow the reaction to proceed further. We know µi (T , p ) for an ideal gas in a mixture and we know that G = ∑ ni µi i ⇒ { } ∆G (ε ) = ε ⎡⎣ν C µC ( g,T , p ) + ν D µD ( g,T , p ) ⎤⎦ − ⎡⎣ν A µA ( g,T , p ) + ν B µB ( g,T , p ) ⎤⎦ 5.60 Spring 2008 Lecture #15 page 2 where ε is an arbitrary small number that allows to let the reaction proceed just a bit. We know that ⎡ pi ⎤ ⎢⎣ 1 bar implied⎥⎦ µi ( g,T , p ) = µio (T ) + RT ln pi where µio (T ) is the standard chemical potential of species “i” at 1 bar and in a pure (not mixed) state. ∴ ⎧⎪ ⎛ pCνC pDνD o o o o ⎡ ⎤ ⎡ ⎤ ∆G (ε ) = ε ⎨ ⎣ν C µC (T ) + ν D µD (T ) ⎦ − ⎣ν A µA (T ) + ν B µB (T ) ⎦ + RT ln ⎜ ν A ν B ⎝ pA pB ⎩⎪ ⇒ ∆G = ∆G ° + RT lnQ ⎞ ⎫⎪ ⎟⎬ ⎠ ⎭⎪ (taking ε=1) where ∆G ° = ⎡⎣ν C µCo (T ) + ν D µDo (T ) ⎤⎦ − ⎡⎣ν A µAo (T ) + ν B µBo (T ) ⎤⎦ and pCνC pCνD Q = νA νB is the reaction quotient pA pB ∆G ° is the standard change in free energy for taking pure reactants into pure products. o o o ∆G o = ∆Grxn = ∆Hrxn −T ∆Srxn or o o ∆G o = ∆Gform ( reactants ) ( products ) − ∆Gform If ∆G (ε ) < 0 then the reaction will proceed spontaneously to form more products ∆G (ε ) > 0 then the backward reaction is spontaneous ∆G (ε ) = 0 No spontaneous changes ⇒ Equilibrium 5.60 Spring 2008 Lecture #15 page 3 o = −RT lnQeq At Equilibrium ∆G (ε ) = 0 and this implies ∆Grxn Define Qeq = K p the equilibrium constant νD νC ⎛ pCνC pCνD ⎞ ∆ν ⎛ XC XC ⎞ K p = ⎜ ν A νB ⎟ = p ⎜ ν A ν B ⎟ = p ∆ν KX ⎝ pA pB ⎠eq ⎝ XA XB ⎠eq and thus o ∆Grxn = −RT ln K p , K p = e −∆G o RT Note from this that K p (T ) is not a function of total pressure p. It is KX = p −∆ν K p which is KX ( p ,T ) . Recall that all pi values are divided by 1 bar, so Kp and KX are both unitless. ________________________________________________ T = 298 K p =1 bar Example: H2(g) + CO2(g) = H2O(g) + CO(g) H2(g) CO2(g) H2O(g) CO(g) Initial # of moles a b 0 0 # moles at Eq. a-x b-x x x Total # moles at Eq. = (a – x) + (b – x) + 2x = a + b Mole fraction a −x a +b b −x a +b x a +b x a +b 5.60 Spring 2008 Lecture #15 page 4 at Eq. o ∆Gform (kJ/mol) ∴ ∆G o rxn and 0 = 28.6 kJ/mol Kp = pH O pCO 2 pH pCO 2 2 = -396.6 Kp = e ⇒ XH OXCO 2 XH XCO 2 2 − -228.6 28,600 kJ/mol (8.314 J/K-mol)(298 K ) -137.2 = e −11.54 = 9.7 x 10 −6 x2 = ( a − x ) (b − x ) Let’s take a = 1 mol and b = 2 mol We need to solve A) x2 (1 − x )(2 − x ) = 9.7 x 10 −6 Using approximation method: K << 1, so we expect x << 1 also. Assume 1 − x ≈ 1, 2 − x ≈ 2 ⇒ x2 (1 − x )(2 − x ) ≈ x2 2 = 9.7 x 10 −6 x ≈ 0.0044 mol (indeed << 1) B) Exactly: x2 = K p = 9.7 x 10 −6 2 x − 3x + 2 x 2 (1 − 9.7x 10 −6 ) + 3x ( 9.7x 10 −6 ) − 2 ( 9.7x 10 −6 ) = 0 x = −3 ( 9.7x 10 −6 ) 2 ( 9.7x 10 −6 ) 9 ( 9.7x 10 −6 ) + 4 (1 − 9.7x 10 −6 ) 2 ( 9.7x 10 −6 ) 2 ± 2 (1 − 9.7x 10 −6 ) The “-“ sign gives a nonphysical result (negative x value) Take the “+” sign only ⇒ x = 0.0044 mol (same) 5.60 Spring 2008 Lecture #15 page 5 Effect of total pressure: example N2O4(g) = 2 NO2(g) Initial mol # n 0 # at Eq. n-x 2x Xi’s at Eq. n −x n +x 2x n +x Total # moles at Eq. = n – x + 2x = n + x 2 Kp = 2 pNO 2 pN O = 2 4 Kp = p 2 p 2XNO pXN O 2 4 4α 2 1 − α2 where α = x n is the fraction reacted K (1 − α ) 4 pp = α 2 2 2 ⎛ 2x ⎞ ⎜ 4x 2 n + x ⎟⎠ =p⎝ =p 2 n −x2 ⎛n − x ⎞ ⎜ ⎟ ⎝n + x ⎠ Kp ⎞ Kp ⎛ α ⎜1 + ⎟= 4 p ⎝ ⎠ 4p 2 Kp 4p 1 = α2 = Kp ⎞ ⎛ ⎛ 4p ⎞ ⎜1 + ⎟ ⎜⎜ 1 + ⎟ 4p ⎠ Kp ⎟ ⎝ ⎝ ∴ ⎛ 4p ⎞ α = ⎜1 + ⎟ ⎜ K p ⎠⎟ ⎝ −1 2 ⎠ If p increases, α decreases Le Chatelier’s Principle, for pressure: An increase in pressure shifts the equilibrium so as to decrease the total # of moles, reducing the volume. In the example above, increasing p shifts the equilibrium toward the reactants. --------------- 5.60 Spring 2008 Lecture #15 page 6 Another example: 2 NO(g) + O2(g) = 2 NO2(g) Initial mol # 2 1 0 # at Eq. 2-2x 1-x 2x Xi’s at Eq. 2 (1 − x ) 3−x 1−x 3−x 2x 3−x Kp = 2 pNO 2 2 pNO pO 2 = 2 p 2XNO = 2 2 p 2XNO pXO 2 2 XNO 1 2 2 p XNO XO 2 = K p = 2.3x 1012 at 298 K Total # moles at Eq. = 2 – 2x + 1 – x + 2x =3-x 2 1 x (3 − x ) p 3 (1 − x ) K p >> 1 so we expect x ≈ 1 ⇒ 3 - x ≈ 2 Kp ≈ 1 2 p (1 − x )3 or (1 − x ) 3 ≈ 2 pK p ⎛ 2 x =1−⎜ ⎜ pK p ⎝ 13 ⎞ ⎟⎟ ⎠ In this case, if p↑ then x↑ as expected from Le Chatelier’s principle.