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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

User Stocked
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1 Answer

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Step-by-step explanation:

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


Kc=([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)=(1)/(Kc)=(1)/(11)=0.091

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


Kc=([H_2SO_4])/([SO_3]^2)

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


Kc_(reverse)=(1)/(Kc)=(1)/(0.0123)=81.3

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


Kc=(1)/([O_2]^3)

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


Kc_(reverse)=(1)/(Kc)=(1)/(1.56)=0.641

User Kabaros
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