Question

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 1

Answer -

Temperature = 25^$\degree$C

K_eq = ?

Some Useful information,

Species	$\Delta$G$\degree$f (kJ/mol)
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) $\rightleftharpoons$ Pb²⁺(aq) + 2 H₂O(l) + Cl₂(g)

We know that,

$\Delta$G^$\degree$ = -RTlnK_eq

K_eq = e^{-$\Delta$G$\degree$/RT} ---------------------A

where,$\Delta$G^$\degree$ = Gibbs Free Energy

R = 8.314 Jmol^-1K^-1

T = Temperature

PbO2(s) + 4 H⁺(aq) + 2 Cl^–(aq) $\rightleftharpoons$ Pb²⁺(aq) + 2 H₂O(l) + Cl₂(g)

Also,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 1 * $\Delta$ G^$\degree$_f(Pb²⁺(aq)) + 2 * $\Delta$ G^$\degree$_f(H₂O(l)) + 1 *$\Delta$G^$\degree$_f(Cl₂(g)) - 1* $\Delta$ G^$\degree$_f(PbO2(s)) - 4 *$\Delta$G^$\degree$_f(H⁺(aq)) - 2* $\Delta$ G^$\degree$_f(Cl^–(aq))

$\Delta$G^$\degree$_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)]

$\Delta$G^$\degree$_rxn = -20.70784 kJ/mol or -20707.84 J/mol [1kJ/mol = 1000J/mol]

Put this in A,

K_eq = e^{-(-20707.84 J/mol/(8.314 Jmol-1K-1 * 298K)}

K_eq = e^8.36

K_eq = 4272.69 or 4.27 * 10^-3 [Answer]

b)

3 O₂(g) + 2 Br^–(aq)  $\rightleftharpoons$ 2 BrO3^–(aq)

We know that,

$\Delta$G^$\degree$ = -RTlnK_eq

K_eq = e^{-$\Delta$G$\degree$/RT} ---------------------A

where,$\Delta$G^$\degree$ = Gibbs Free Energy

R = 8.314 Jmol^-1K^-1

T = Temperature

3 O₂(g) + 2 Br^–(aq)  $\rightleftharpoons$ 2 BrO3^–(aq)

Also,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 2 * $\Delta$ G^$\degree$_f(BrO3^–(aq)) - 3* $\Delta$ G^$\degree$_f(O₂(g)) - 2 *$\Delta$G^$\degree$_f(Br^–(aq))

$\Delta$G^$\degree$_rxn = [2 * (1.6736 kJ/mol)] - [3 * (0 kJ/mol)] - [2 * (-103.9724 kJ/mol)]

$\Delta$G^$\degree$_rxn = 211.292 kJ/mol or 211292 J/mol [1kJ/mol = 1000J/mol]

Put this in A,

K_eq = e^{-(211292 J/mol/(8.314 Jmol-1K-1 * 298K)}

K_eq = e^85.28

K_eq = 1.09 * 10³⁷ [Answer]