Given the following equilibrium constants at 431°C,

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Given the following equilibrium constants at 431°C,
Na2 0(s) = 2 Na(1) + 02 (g) K1 = 7× 10-25
2
1
O2 (g)
NaO(g) = Na(l) +
K2 = 3 x 10-5
Na2 O2 (s) = 2 Na(1) + O2 (9)
K3 = 3 x 10-29
NaO2 (s) = Na(1) + O2 (9)
K4 = 4 x 10-14
determine the values for the equilibrium constants for the following reactions:
a. Na2 0(s) + 02 (g) = Na2 O2 (s)
2
Equilibrium constant =
b. NaO(g) + Na2 0(s) = Na2 O2 (s) + Na(1)
Equilibrium constant =
c. 2 NaO(g) = Na2 O2 (s)
Equilibrium constant =
(Hint: When reaction equations are added, the equilibrium expressions are multiplied.)
Transcribed Image Text:Given the following equilibrium constants at 431°C, Na2 0(s) = 2 Na(1) + 02 (g) K1 = 7× 10-25 2 1 O2 (g) NaO(g) = Na(l) + K2 = 3 x 10-5 Na2 O2 (s) = 2 Na(1) + O2 (9) K3 = 3 x 10-29 NaO2 (s) = Na(1) + O2 (9) K4 = 4 x 10-14 determine the values for the equilibrium constants for the following reactions: a. Na2 0(s) + 02 (g) = Na2 O2 (s) 2 Equilibrium constant = b. NaO(g) + Na2 0(s) = Na2 O2 (s) + Na(1) Equilibrium constant = c. 2 NaO(g) = Na2 O2 (s) Equilibrium constant = (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)
Expert Solution
Step 1 introduction

 

Given equilibrium reactions are 

1. Na2O (s)  <--> 2 Na (l)  + 1/2 O2 (g)                K1 = 7 x 10-25      

2. NaO (g)  <--> Na (l)  + 1/2 O2 (g)                   K2 = 3 x 10-5  

3. Na2O2(s) <--> 2 Na(l) + O2 (g)                        K3 = 3 x10-29 

4.  NaO2 (s) <--> Na(l)  + O2 (g)                           K= 4 x10-14 

 

If reactions added , the equilibrium expressions multiplied , if reaction is revesed K become reciprocated 

 

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