8.2. Chemical reactions[[Student version, January 17, 2003]] 267
- Atlow temperatures, the first factor becomes extremely large, because it is the exponential
of a large positive number. Equation 8.11 in turn implies that the equilibrium shifts almost
completely in favor of water. - Atvery high temperatures the first factor is close to 1. Equation 8.11 then says that there
will be significant amounts of unreacted hydrogen and oxygen. The mechanical interpretation
is that thermal collisions are constantly ripping water molecules apart as fast as they form.
Example Physical chemistry books give the equilibrium constant for Reaction 8.7 at room
temperature as e(457kJ/mole)/kBTr.Ifwebegin with exactly two moles of hydrogen
and one of oxygen in a 22Lchamber, how much of these reactants will remain after
the reaction comes to equilibrium?
Solution: Weuse Equation 8.11. Recalling that standard free energy changes for
gases are computed using the standard concentration 1 mole/ 22 L,and lettingxbe
the number of moles of unreacted O 2 ,wehaveinthe final state 2(1−x)moles of
H 2 O, and hence 2xmoles of unreacted H 2 .ThusEquation 8.11 says
[2(1−x)]^2
(2x)^2 (x)(1 mole/ 22 L)
=e(457kJ/mole)/(2.^5 kJ/mole)22 L
mole.
Almost all of the reactants will be used up, because the reaction is energetically
very favorable. Thusxwill be very small and we may approximate the numerator
on the left, replacing (1−x)by1.The amount of oxygen left is then the reciprocal
of the cube root of the quantity on the right, orx=3. 4 · 10 −^27 .There is twice as
muchhydrogen left, or 6. 8 · 10 −^27 mole.General reactions Quite generally we can consider a reaction amongkreactants andm−k
products:
ν 1 X 1 +···+νkXkνk+1Xk+1+···+νmXm.
The whole numbersνkare called thestoichiometric coefficientsof the reaction. Defining
∆G≡−ν 1 μ 1 −···−νkμk+νk+1μk+1+···+νmμm, (8.13)weagain find that−∆Gis the free energy change when the reaction takes one forward step, or:
Achemical reaction will run forward if the quantity∆Gis a negative number,
or backward if it’s positive.(8.14)
Idea 8.14 justifies us in calling ∆Gthe netchemical forcedriving the reaction. Equilibrium is
the situation where a reaction makes no net progress in either direction, or ∆G=0.Just as before,
wecan usefully rephrase this condition by separating ∆Ginto its concentration-independent part,
∆G^0 ≡−ν 1 μ^01 −···+νmμ^0 m, (8.15)plus the concentration terms. Defining theμ^0 ’s with standard concentrationsc 0 =1Mgives the
general form of Equation 8.11, called theMass Actionrule:
[Xk+1]νk+1···[Xm]νm
[X 1 ]ν^1 ···[Xk]νk
=Keq in equilibrium, where Keq≡e−∆G(^0) /kBT
. (dilute limit)
(8.16)