lecture 23 - Chemical equilibrium continued: the...

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Unformatted text preview: Chemical equilibrium continued: the general approach Extent of reaction First of all, we will write any chemical reaction in the following general form: ∑ν A = 0 i i i Here νi‘s are the stoichiometric coefficients, which are taken negative for reactants (whatever is on the lhs) and positive for products (rhs). For example: N2 + 3H2 = 2NH3 νi  ­1  ­3 2 Ai N2 H2 NH3 Let n(A) be the number of moles of compound A in the mixture. If say 1 mole of N2 is converted in the reaction, this would require 3 moles of H2. We see that dn(N2)/( ­1) = dn(H2)/( ­3) = dn(NH3)/2 Generally, we will have dn1/ν1 = dn2/ν2 = … = dξ, where we have introduced a new variable, ξ, called the extent of reaction. If we additionally require that ξ=0 at the beginning of the experiment, then we get ni = ξνi + ni(0), where ni(0) is the initial number of moles of compound i. Example. Suppose that, for the above reaction, we initially have 1 mole of H2 and 1 mole of N2. Suppose also that at some later moment the number of moles of nitrogen was measured to be (a) n(N2) = 2/3 and (b) ½. What are the amounts of the other two components in the system? Solution. (a)We have n(N2) = 1  ­ξ = 2/3. Therefore ξ=1/3. n(H2) = 1 ­3ξ=0. n(NH3)=2ξ = 2/3. (b) In a similar way, we find ξ=1/2. This gives n(H2) = 1 ­3ξ =  ­1/2. This is an absurd value since the number of moles must be positive. Therefore the amount of nitrogen in this system cannot be equal to ½. The equilibrium condition. For a mixture that consists of the reactants and the products we can write, at constant P and T: dG = ∑ µ dn i i In equilibrium we have dG=0. Using dni = dξνi, we conclude ∑µν = 0 i i Assuming ideal mixture and writing the chemical potential of each component as µi = µi0 + RT ln xi , after some manipulations we find ∏ xν = exp( ­ΔG /RT) = K i i 0 where K is equilibrium constant and ΔG0 = ∑µ ν 0 ii Example 1. For the reaction O2=2O we find: x(O)2/x(O2) = exp( ­ΔG0(P,T)/RT), as previously derived. ...
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