Determine the numerical value of Keq at 25 °C for the electrochemical reactions below. I haven’t lectured on this in class, but the “how-to” is clearly illustrated in the textbook in Chapter 19. (a) PbO2(s) + 4 H+(aq) + 2 Cl–(aq) → ← Pb2+(aq) + 2 H2O(l) + Cl2(g) (acidic solution) Keq____________________________ (b) 3 O2(g) + 2 Br–(aq) → ← 2 BrO3–(aq) (basic solution) Keq____________________________
Answer -
Temperature = 25C
Keq = ?
Some Useful information,
Species | ![]() ![]() |
---|---|
PbO2(s) | -215.476 |
H+(aq) | 0 |
Cl–(aq) | -131.25208 |
Pb2+(aq) | -24.43 |
H2O(l) | -237.178408 |
Cl2(g) | 0 |
O2(g) | 0 |
Br–(aq) | -103.9724 |
BrO3–(aq) | 1.6736 |
a)
PbO2(s) + 4 H+(aq) + 2
Cl–(aq)
Pb2+(aq) + 2 H2O(l) + Cl2(g)
We know that,
G
= -RTlnKeq
Keq =
e-G
/RT
---------------------A
where,G
= Gibbs Free Energy
R = 8.314 Jmol-1K-1
T = Temperature
PbO2(s) + 4 H+(aq) + 2
Cl–(aq)
Pb2+(aq) + 2 H2O(l) + Cl2(g)
Also,
G
rxn
=
n *
G
f(products)
-
n *
G
f(reactants)
where, n = stiochiometric coefficient
G
rxn
= 1 *
G
f(Pb2+(aq))
+ 2 *
G
f(H2O(l))
+ 1 *
G
f(Cl2(g))
- 1*
G
f(PbO2(s))
- 4 *
G
f(H+(aq))
- 2*
G
f(Cl–(aq))
G
rxn
= [1 * (-24.43 kJ/mol)] + [2 * (-237.129 kJ/mol)] + [1 * (0
kJ/mol)] - [1* (-215.476 kJ/mol)] - [4 * (0 kJ/mol)] - [2 *
(-131.25208 kJ/mol)]
G
rxn
= -20.70784 kJ/mol or -20707.84
J/mol [1kJ/mol = 1000J/mol]
Put this in A,
Keq = e-(-20707.84 J/mol/(8.314 Jmol-1K-1 * 298K)
Keq = e8.36
Keq = 4272.69 or 4.27 * 10-3 [Answer]
b)
3 O2(g) + 2
Br–(aq)
2 BrO3–(aq)
We know that,
G
= -RTlnKeq
Keq =
e-G
/RT
---------------------A
where,G
= Gibbs Free Energy
R = 8.314 Jmol-1K-1
T = Temperature
3 O2(g) + 2
Br–(aq)
2 BrO3–(aq)
Also,
G
rxn
=
n *
G
f(products)
-
n *
G
f(reactants)
where, n = stiochiometric coefficient
G
rxn
= 2 *
G
f(BrO3–(aq))
- 3*
G
f(O2(g))
- 2 *
G
f(Br–(aq))
G
rxn
= [2 * (1.6736 kJ/mol)] - [3 * (0 kJ/mol)] - [2 * (-103.9724
kJ/mol)]
G
rxn
= 211.292 kJ/mol or 211292 J/mol [1kJ/mol =
1000J/mol]
Put this in A,
Keq = e-(211292 J/mol/(8.314 Jmol-1K-1 * 298K)
Keq = e85.28
Keq = 1.09 * 1037 [Answer]
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