CHEM1901/3 Worksheet 11: Free Energy And Equilibrium Model ...€¦ · CHEM1901/3 Worksheet 11:...
5
CHEM1901/3 Worksheet 11: Free Energy And Equilibrium Model 1: The Equilibrium Constant As discussed in Model 1 of week 10, nitrogen dioxide, NO 2 , is a prominent air pollutant which is in equilibrium with its dimer, N 2 O 4 . Starting from NO 2 , the formation of the dimer can be studied using either equation (A) or (B) below. Starting from the dimer, the formation of NO 2 can be studied using (C) or (D). 2NO 2 (g) N 2 O 4 (g) (A) N 2 O 4 (g) 2NO 2 (g) (C) NO 2 (g) ½ N 2 O 4 (g) (B) ½ N 2 O 4 (g ) NO 2 (g) (D) Many chemical reactions lead to a mixture of reactants and products. You will end up with a mixture of both NO 2 (g) and N 2 O 4 (g) whether you start with pure NO 2 (g) or pure N 2 O 4 (g). Such reactions are said to reach an equilibrium in which the amount of each substance does not change. Consider a reaction such as that below which has been left long enough to reach equilibrium. wW(g) + xX(g) yY(g) + zZ(g) The equilibrium constant in terms of concentrations, K c , is a constant at a given temperature that defines how much of each substance there will be at equilibrium: K c = [! ! ] ! [! ! ] ! [! ! ] ! [! ! ] ! If K c > 1, the mixture will contain more of the substances on the right hand side (Y and Z) of the equation. If K c < 1, the mixture will contain more of the substances on the left hand side (W and Z) of the equation. Critical thinking questions 1. Write down the expression for K c for reactions A, B, C and D in Model 1. K c (A) = K c (B) = K c (C) = K c (D) = 2. Looking at the equations in Q1, what is the mathematical relationship between the different forms of K c ? (a) K c (A) and K c (B) (b) K c (A) and K c (C) 3. At equilibrium at room temperature, [NO 2 (g)] = 1.60 M and [N 2 O 4 ] = 0.20 M. Calculate the values of K c (A), K c (B) and K c (C) and hence confirm your analysis in Q2. Model 2: The Reaction Quotient The reaction quotient, Q c , for a reaction wW(g) + xX(g) yY(g) + zZ(g), is defined as follows: Q c = [! ! ] ! [! ! ] ! [! ! ] ! [! ! ] ! It looks similar to the equilibrium constant expression. The difference is that Q c can be calculated at any time during a reaction or if a reaction is disturbed. It is used to predict the direction in which a reaction will move.
Transcript of CHEM1901/3 Worksheet 11: Free Energy And Equilibrium Model ...€¦ · CHEM1901/3 Worksheet 11:...
CHEM1901/3 Worksheet 11: Free Energy And Equilibrium Model 1: The Equilibrium Constant As discussed in Model 1 of week 10, nitrogen dioxide, NO2, is a prominent air pollutant which is in equilibrium with its dimer, N2O4. Starting from NO2, the formation of the dimer can be studied using either equation (A) or (B) below. Starting from the dimer, the formation of NO2 can be studied using (C) or (D).
2NO2(g) N2O4(g) (A) N2O4(g) 2NO2(g) (C) NO2(g) ½ N2O4(g) (B) ½ N2O4(g ) NO2(g) (D)
Many chemical reactions lead to a mixture of reactants and products. You will end up with a mixture of both NO2(g) and N2O4(g) whether you start with pure NO2(g) or pure N2O4(g). Such reactions are said to reach an equilibrium in which the amount of each substance does not change. Consider a reaction such as that below which has been left long enough to reach equilibrium.
wW(g) + xX(g) yY(g) + zZ(g) The equilibrium constant in terms of concentrations, Kc, is a constant at a given temperature that defines how much of each substance there will be at equilibrium:
Kc =[! ! ]![! ! ]!
[! ! ]![! ! ]!
If Kc > 1, the mixture will contain more of the substances on the right hand side (Y and Z) of the equation. If Kc < 1, the mixture will contain more of the substances on the left hand side (W and Z) of the equation.
Critical thinking questions 1. Write down the expression for Kc for reactions A, B, C and D in Model 1.
Kc (A) = Kc (B) =
Kc (C) = Kc (D) =
2. Looking at the equations in Q1, what is the mathematical relationship between the different forms of Kc?
(a) Kc (A) and Kc (B) (b) Kc (A) and Kc (C)
3. At equilibrium at room temperature, [NO2(g)] = 1.60 M and [N2O4] = 0.20 M. Calculate the values of Kc(A), Kc(B) and Kc(C) and hence confirm your analysis in Q2.
Model 2: The Reaction Quotient The reaction quotient, Qc, for a reaction wW(g) + xX(g) yY(g) + zZ(g), is defined as follows:
Qc =[! ! ]![! ! ]!
[! ! ]![! ! ]!
It looks similar to the equilibrium constant expression. The difference is that Qc can be calculated at any time during a reaction or if a reaction is disturbed. It is used to predict the direction in which a reaction will move.
From model 1, at equilibrium [NO2(g)] = 1.60 M, [N2O4] = 0.20 M and Kc = 0.078 for reaction (A):
Critical thinking questions 1. Predict what will happen to this reaction if more NO2 is added so that [NO2(g)] = 2.00 M?
2. Predict what will happen to this reaction if instead NO2 is removed so that [NO2(g)] = 1.00 M?
3. Calculate the values for Qc for these two experiments.
(a) [NO2(g)] = 2.00 M and [N2O4] = 0.20 M: Qc =
(b) [NO2(g)] = 1.00 M and [N2O4] = 0.20 M: Qc =
4. Using your answers to Q1 - 3, what happens to a reaction if
(a) Qc < Kc
(b) Qc > Kc
5. The equilibrium constant for the dissolution of silver iodide at 25 °C is 1.5 × 10–16. Calculate the equilibrium concentrations of Ag+(aq) and I–(aq) if 0.200 mol of AgI(s) were dispersed in:
(a) 1.0 L of water
(b) 1.0 L of a 0.0050 M aqueous solution of KI.
Model 3: Ellingham Diagrams Ellingham diagrams are plots of the change in the equilibrium constant (which is related to the standard free energy change by ΔrxnGo = -RTlnKp) as a function of temperature. They are particularly useful for determining under what conditions a metal ore will be reduced to pure metal. Look at the simplified, single system Ellingham diagram overleaf.
Equilibrium
The values of lnKp correspond to the two reactions below:
Zn(s) + ½ O2(g) à ZnO(s) (A) C(s) + ½ O2(g) à CO(g) (B) To extract Zn from ZnO using C, the reaction below needs to be favourable:
ZnO(s) + C(s) à Zn(s) + CO(g) (C)
Critical thinking questions 1. Write down the form of Kp for reactions (A), (B) and (C)
Kp (A) = Kp (B) = Kp (C) =
2. Write down the form of lnKp for reactions (A), (B) and (C)
lnKp (A) = lnKp (B) = lnKp (C) =
3. What is the mathematical relationship between Kp (A), Kp (B) and Kp (C) and between lnKp (A), lnKp (B) and lnKp (C)?
4. At 500 K, the Zn-ZnO line has higher value of lnKp than the C-CO line. Using your answer to Q2,
explain what this means for the value of the equilibrium constant for the extraction reaction (C).
5. Above 1050 K, the Zn-ZnO line is below the C-CO line. Using your answer to Q2, explain what this
means for the value of the equilibrium constant for the extraction reaction (C).
0
20
40
60
80
100
400 600 800 1000 1200 1400 1600
ln K
p
Temperature / K
Zn-‐ZnO
C-‐CO
6. At ~1050 K, the Zn-ZnO line crosses the C-CO line. Using your answer to Q2, explain what this means for the value of the equilibrium constant for the extraction reaction (C).
The questions below correspond to the past exam questions 2009-J-12 and 2009-J-13.
7. Write the equations for the combustion of graphite that occur in smelting processes to produce (i) CO(g) and (ii) CO2(g), in which one mole of O2(g) is consumed.
8. Use the standard enthalpy and entropy data provided to calculate the enthalpy and entropy of these two combustion reactions of graphite.
substance ΔfH° / kJ mol–1 S° / J K–1 mol–1 substance S° / J K–1 mol–1
CO(g) –111 198 C(s) 6
CO2(g) –394 214 O2(g) 205
Equilibrium
9. Using calculations where necessary, sketch the temperature-dependence of the equilibrium constant for these two combustion reactions on the graph shown below. (Hint: remember that ΔrxnGo = -RTlnKp and ΔrxnGo = ΔrxnHo -TΔrxnSo.)
10. Over what temperature range is carbon monoxide the favoured product?
