Topic 4.3 Chemical equilibrium

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4.3 (a) Partial pressure and concentration
Concentration is the amount of a substance in a given volume.  For example aqueous sodium hydroxide solution having a concentration of 2moldm^-3.  This is represented as [NaOH(aq)] = 2moldm^-3.  Note [] indicates a concentration.
Partial pressure is the contribution to the total pressure in a gas mixture made by just one of the components in the mixture when it is at equilibrium.  The partial pressure of a component A in an equilibrium mixture has the symbol p_A.
Total pressure of gas in a mixture = sum of all partial pressures.
For the equilibrium H₂(g) + I₂(g) <=> 2HI(g)
P_total = pHI + pH₂ + pI₂
Partial pressure = total pressure * mole fraction
p_HI = P_total * c_HI
c_HI = amount of HI / total amount of HI, H₂ and I₂

4.3 (b) Expressions for Kc and Kp
For the equation
aP + bQ <=> cR + dS
the equilibrium constant Kc is:           Kc = [R]^c [S]^d /[P]^a [Q]^b
(Square brackets represent concentrations in mol dm^-3)

Example: Calculate Kc for the esterification of ethanoic acid by ethanol given that for a 1dm3 of this homogeneous liquid equilibrium the amounts present are as shown below.
                                CH₃CO₂H + C₂H₅OH <=>CH₃CO₂C₂H₅ + H₂O
Equilibrium amount/mol 0.0255     0.0245           0.0584              0.0457
K_c = [CH₃CO₂C₂H₅] [H₂O] / [CH₃CO₂H] [C₂H₅OH]
K_c = 0.0584 x 0.0437/ 0.0255 x 0.0245 = 4.1 (no units for esterification reaction)
For the equilibrium H₂(g) + I₂(g) <=> 2HI(g)
The equilibrium constant K_p for this reaction in terms of partial pressures is:
K_p = p²HI / pH₂*pI₂
units = atm²/atm*atm = no units for this K_p
In general for equilibrium
aA(g) + bB(g) = cC(g) + dD(g)
the equilibrium constant is,
K_p = p^c_C p^d_D/ p^a_A p^b_B
Example: If at 55^oC the partial pressure of nitrogen dioxide in an equilibrium mixture is 0.67atm and the partial pressure of dinitrogen tetraoxide in the mixture is 0.33atm what is the value of K_p for the reaction at this temperature?
N₂O₄(g) <--> 2NO₂(g)
Kp = p²NO₂(g)/ pN₂O₄(g)
Kp = 0.67atm²/0.33atm
Kp = 1.36atm
Task 4.3b This question is about the equilibrium system:
                                CH₃CO₂H + C₂H₅OH <=>CH₃CO₂C₂H₅ + H₂O
In an experiment, 0.33 mol of CH₃CO₂H, 0.66 mol of CH₃CO₂C₂H₅  and 0.66 mol of H₂O was found to be present at equilibrium.  What amount of C₂H₅OH is present?  answer

4.3 (c) Values for solid and liquid phase in Kc and Kp
Some equilibrium reactions involve only substances in the same phase e.g. all aqueous solutions or all gases.  These are homogeneous reactions.
Some equilibrium reactions involve substances in different phases e.g. a pure liquid and a gas or a solid and a gas.  These are heterogeneous reactions.  Expressions for K_c and K_p do not include values for solid and pure liquid phases in heterogeneous reactions.
For example, the heterogeneous reaction between iron and steam
3Fe (s) + 4H₂O (g) = Fe₃O₄ (s) + 4H₂ (g)
Kp = p⁴H₂ / p⁴H₂O
Solids do not appear in the expression because their vapour pressures remain constant ( at a constant temperature) as long as there is some of each solid present (same applies for liquids) These constant vapour pressures are incorporated into the value of K_p .

4.3 (d) Catalyst and the position of equilibrium
Catalysts do not alter the equilibrium constant (K_p or K_c) or the position of equilibrium. They do affect the time needed for the system to reach equilibrium.

4.3 (e) Finding equilibrium partial pressures given Kp values
Example: What is the partial pressure of nitrogen dioxide in an equilibrium mixture if the partial pressure of dinitrogen tetraoxide in the mixture is 0.33atm and K_p at 55^oC is 1.36atm for the reaction
N₂O₄(g) = 2NO₂(g)
Kp = p²NO₂(g)/ pN₂O₄(g)
p²NO₂(g) = Kp * pN₂O₄(g) = 1.36atm * 0.33atm = 0.45atm
pNO₂(g) = 0.67atm

4.3 (f) The effect of temperature on Kc and Kp
K_c, K_p and the position of equilibrium are affected by temperature in endothermic and exothermic equilibria.  The effects are the same as predicted by Le Chatiers principle.
Exothermic reaction:
Temperature rise:- position of equilibrium moves to left, K_p and K_c become smaller.
Temperature fall:-  position of equilibrium moves to right, K_p and K_c become bigger
Endothermic reaction:
Temperature rise:- position of equilibrium moves to right, K_p and K_c become bigger.
Temperature fall:-  position of equilibrium moves to left, K_p and K_c become smaller.