lecture 24

# lecture 24 - Chemical equilibrium continued We have...

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Unformatted text preview: Chemical equilibrium continued We have derived the general relationship describing chemical equilibrium for any chemical reaction: ∏ xν = exp( ­ΔG /RT) = K i i 0 where K is equilibrium constant and ΔG0 = ! µ ! = ! "G 0 ii f ,i ! i Example 2. For the reaction N2 + 3H2 = 2NH3 0 x 2 ( NH 3 ) = e−ΔG / RT , 3 x( N 2 ) x ( H 2 ) where ΔG0=2µ0(NH3)  ­ µ0(N2) ­ 3µ0(H2) = 2ΔGf(NH3)  ­ ΔGf (N2) ­ 3 ΔGf (H2). For example, at room T and P=1 atm ΔG0 = 2 ( ­16.367) – 0 – 0 =  ­32.734 kJ/mol, giving the equilibrium constant K ~ exp(13) . The large value of K seems to suggest that if we mix hydrogen and nitrogen at room temperature and atmospheric pressure, they will be easily converted into NH3. This does not happen. In industry, ammonia has to be synthesized under high pressure and at high temperature. In addition, a catalyst is needed. The equilibrium constant tells us about the equilibrium but does not tell us how fast it is reached. It may take more than the age of the Universe to form products for a particular chemical reaction yet these products may be thermodynamically favored. How fast the chemical reaction takes place is an issue studied by kinetics. We will address it later. Example 3. Finally, let us put all we have learned together and calculate the equilibrium composition for the reaction N2 + 3H2 = 2NH3 given that the initial amounts of NH3, H2, and N2 are 0.1 mole, 3 mole, and 1 mole, respectively. Solution. n(NH3) = 0.1 + 2ξ n(H2) = 3 ­3ξ n(N2) = 1 ­ξ ntotal = n(NH3) + n(H2)+ n(N2) = 4.1 ­2ξ For the mole fractions, we find x(NH3) = n(NH3)/ntotal = (0.1 + 2ξ)/(4.1 ­2ξ) x(H2) = (3 ­3ξ)/(4.1 ­2ξ) x(N2) = (1 ­ξ)/(4.1 ­2ξ) The equilibrium condition is: K = x2(NH3)/(( x3(H2) x(N2))) = (0.1 + 2ξ)2 (4.1 ­2ξ)2/((3 ­3ξ)3(1−ξ)) After the equilibrium constant K is calculated, the value of ξ can be found by solving this equation. Suppose, for instance, that at some temperature and pressure the value of K was found to be K = 1. Then we have the equation (0.1 + 2ξ)2 (4.1 ­2ξ)2=(3 ­3ξ)3(1−ξ). This kind of equation is generally hard to solve analytically but can be easily solved on a computer. We know from calculus that it should generally have 4 solutions. Which one is the correct one? Solving this equation numerically, I found that two roots are complex (i.e., of the form a+b i) and therefore unphysical. The amount of each material is a real number and so is the extent of reaction. The remaining two roots are ξ=0.3075 and ξ=1.692. If we choose the second of them, this leads to x(N2) =  ­0.969, x(NH3) = 4.874, x(H2) =  ­2.905. These values are absurd because the mole fractions must be positive and in the range between 0 and 1. The only remaining solution, ξ=0.3075, gives us x(N2) = 0.199, x(NH3) = 0.596, x(H2) = 0.295. The lesson here is that when solving for the extent of reaction, one may find several different solutions. The ones that do not satisfy physical constraints (that mole fractions must be real numbers in the range from 0 to 1) should be dropped. K(T) and K(P) We have ln(K) =  ­ ∑µ ν 0 ii /RT Differentiating this with respect to T, we get ∂ ln K / ∂T = −(1/ R)∑ν i ∂(µio / T ) / ∂T For the chemical potential, we have: ∂ ⎛ µ ⎞ 1 ∂µ µ − S µ H − 2= − 2 = − 2 ⎜ ⎟= ∂T ⎝ T ⎠ T ∂T T TT T so that ∂ ln K / ∂T = (1/ R)∑ν i ( Hio / T 2 ) = ΔH / RT 2 where ΔH is the heat of reaction. This is the famous van’t Hoff equation. If we know how ΔH depends on T then we can calculate K(T) at constant pressure. ln K ( P, T ) / K ( P, T0 ) = ∫ dT ′ΔH (T ′) / RT ′2 T0 T We have discussed how the temperature dependence of ΔH(T) can be found if one knows the dependence of the heat capacities on temperature: ΔH (T ′) = ΔH (T0 ) + ∑ i T′ T0 ∫ν C i i p (T ′′)dT ′′ Now fix T and consider K(P). To find K(P), we differentiate the expression for lnK with respect to P: ∂ ln K / ∂P = − ∑ (∂µ 0 i / ∂P)ν i / RT = − ∑ Vi0 νi / RT = −ΔV / RT Here ΔV is the change of the volume in going from the reactants to the products. If ALL reactants and products are ideal gases then Vi0 = RT / P so that ΔV = ( RT / P)∑ν i Integration then gives ln[K(P,T)/K(P0,T)] = −( Examples: For the reaction N2 (g) + 3H2 (g) = 2NH3(g), ΔV =  ­2RT/P <0. This means that the equilibrium constant will increase with the increasing pressure, ln[K(P,T)/K(P0,T)] = 2 ln(P/P0), or K ( P, T ) = K ( P0 , T )( P / P0 )2 . For the reaction O2 = 2O we find that the volume increases: this means that K(P) decreases with increasing pressure (as previously discussed). Specifically, we find ∑ ν )ln(P / P ) i 0 K ( P, T ) = K ( P0 , T )( P / P0 )!1 ...
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## This note was uploaded on 04/24/2011 for the course CH 52635 taught by Professor Makarov during the Spring '11 term at University of Texas.

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