1. 2008-09 NSERC Undergraduate Student Research Awards (USRA)
Department deadline is Friday, February 1, 2008
In addition to the research experience, you will earn $ for 16 weeks + at least $1, from your professor.
Students must have obtained a cumulative GPA of at least 8.0 or B- up to the time of application

2. 59-241 Labs Find your lab partner within the next few days.
TAs will be in the classroom next Tuesday to sign you up and may also tell you which experiment you will start with
Tutorial time is now posted on the website.

3. Expressions of the equilibrium constant K
The equilibrium constant, K, (a dimensionless quantity) can be expressed in terms of fugacities for gas phase reactions or activities for aqueous phase reactions.
Fugacity ( a dimensionless quantity) is equal to the numerical value of partial pressure, i.e. pj/pθ where pθ = 1 bar).
The activity, a, is equal to the numerical value of the molality, i.e. bj/bθ where bθ = 1 mol kg-1.

4. For reactions occurring in electrolyte solutions
Effects of the interactions of ions on the reaction process should be considered.
Such a factor can be expressed with the activity coefficient, γ, which denotes distance from the ideal system where there is no ion-interaction.
The activity shall now be calculated as αj = γj*bj/bθ
For a reaction A + B ↔ C + D
K =

5. The activities of solids are equal to 1
Illustration: Express the equilibrium constant for the heterogeneous reaction
NH4Cl(s) ↔ NH3(g) + HCl(g)
Solution:
In term of fugacity (i.e. partial pressure): Kp =
In term of molar fraction: Kx =

6. Estimate reaction compositions at equilibrium
Example 1: Given the standard Gibbs energy of reaction H2O(g) → H2(g) + 1/2O2(g) at 2000K is kJ mol-1, suppose that steam at 200k pa is passed through a furnace tube at that temperature. Calculate the mole fraction of O2 present in the output gas stream.
Solution: (details will be discussed in class)
lnK = - (135.2 x 103 J mol-1)/( JK-1mol-1 x 2000K) =
K = x10-4
K =
Ptotal = 200Kpa
assuming the mole fraction of O2 equals x
PO2 = x* Ptotal,
PH2 = 2(x*Ptotal)
PH2O = Ptotal – PO2 – PH2 = (1-3x)Ptotal

7. Equilibria in biological systems
Biological standard state: pH = 7.
For a reaction: A + vH+(aq) ↔ P
ΔrG = ΔrGθ + RT
= ΔrGθ + RT
the first two terms of the above eq. form ΔrG‡
ΔrG‡ = ΔrGθ vRTln10

8. Solution: ΔrG‡ = ΔrGθ + 7vRTln10
Example: For a particular reaction of the form A → B + 2H+ in aqueous solution, it was found that ΔrGθ = 20kJ mol-1 at 28oC. Estimate the value of ΔrG‡.
Solution: ΔrG‡ = ΔrGθ vRTln10
here v = - 2 !!!
ΔrG‡ = 20 kJ mol (-2)(8.3145x10-3 kJ K-1mol-1)
x( K)ln10
= 20 kJ mol-1 – kJ mol-1
= -61 kJ mol-1
(Note that when measured with the biological standard, the standard reaction Gibbs energy becomes negative!)

9. Molecular Interpretation of equilibrium

10. The response of equilibria to reaction conditions
Equilibria respond to changes in pressure, temperature, and concentrations of reactants and products.
The equilibrium constant is not affected by the presence of a catalyst.

11. How equilibria respond to pressure
Equilibrium constant K is a function of the standard reaction Gibbs energy, ΔrGθ .
Standard reaction Gibbs energy ΔrGθ is defined at a single standard pressure and thus is independent of pressure.
The equilibrium constant is therefore independent of pressure:

12. consider the reaction 2A(g) ↔ B(g)
K is independent of pressure does NOT mean that the equilibrium composition is independent of the pressure!!!
consider the reaction 2A(g) ↔ B(g)
assuming that the mole fraction of A equals xA at quilibrium, then xB = 1.0 – xA,
K =
because K does not change, xA must change in response to any variation in Ptotal!!!

13. Le Chatelier’s Principle
A system at equilibrium, when subject to a disturbance, responds in a way that tends to minimize the effect of the disturbance.

14. Example: Predict the effect of an increase in pressure on the Haber reaction, 3H2(g) + N2(g) ↔ 2NH3(g).
Solution:
According to Le Chatelier’s Principle, an increase in pressure will favor the product.
prove: K =
Therefore, to keep K unchanged, the equilibrium mole fractions Kx will change by a factor of 4 if doubling the pressure ptotal.

15. The response of equilibria to temperature
According to Le Chatelier’s Principle:
Exothermic reactions: increased temperature favors the reactants.
Endothermic reactions: increased temperature favors the products.
The van’t Hoff equation:
(a) (7.23a)
(b) (7.23b)

16. Derivation of the van’t Hoff equation:
Differentiate lnK with respect to temperature
Using Gibbs-Helmholtz equation (eqn th edition)
thus
Because d(1/T)/dT = -1/T2:

17. For an exothermic reaction, ΔrHθ < 0, thus , suggesting that increasing the reaction temperature will reduce the equilibrium constant.