Question

Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc for the reverse reaction.
(1) Al(s) + NaOH(aq) + H2O(l) ⇋ Na[Al(OH)4](aq) + H2(g) Kc for balanced reaction = 11
(2) H2O(l) + SO3(g) ⇋ H2SO4 (aq) Kc for balanced reaction = 0.0123
(3) P4(s) + O2(g) ⇋ P4O6(s) Kc for balanced reaction = 1.56

Answer 1

Explanation:

1) $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)$

$Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091$

2) $H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)$

$Kc=\frac{[H_2SO_4]}{[SO_3]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3$

3)  $P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)$

$Kc=\frac{1}{[O_2]^3}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641$