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please help five munite .give the following equilibrium constat at 427°C:

Chemistry
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O2(g) + 4Na(l)  2Na2O(s) K1 = 2.50×1049

O2(g) + 2Na(l)  2NaO(g) K2 = 2.50×109

O2(g) + 2Na(l)  Na2O2(s)  K3 = 2.00×1028

O2(g) + Na(l)  NaO2(s) K4 = 3.33×1013

Wht would,be value equibrium constat each reactions,at 427°C?

Na2O2(s) +Na(l)  NaO(g)+Na2O(s)


Feb 1st, 2015

Given the following equilibrium constants at 427°C: 
O2(g) + 4Na(l) 2Na2O(s) K1 = 2.50×1049 
O2(g) + 2Na(l) 2NaO(g) K2 = 2.50×109 
O2(g) + 2Na(l) Na2O2(s) K3 = 2.00×1028 
O2(g) + Na(l) NaO2(s) K4 = 3.33×1013 
What would be the value of the equilibrium constant for each of the following reactions, at 427°C? 
Na2O2(s) + O2(g) 2NaO2(s) 
5.54×10-2 
The Keq for a set of reactions is the Keq of each reaction multiplied by one another. To get the given equation, I would have to add –reaction 3 to 2x reaction 4:observe 

-1{O2(g) + 2Na(l) Na2O2(s)} 
2{ O2(g) + Na(l) NaO2(s)} 

Na2O2(s)  O2(g) + 2Na(l) 
+2O2(g) + 2Na(l) 2NaO2(s) 
Na2O2(s) + O2(g) 2NaO2(s) 

The reversing of equation 3 affects K3. Since K is defined as K forward/K backwards 
1/K3 = K3 for the reverse reaction 
The doubling of equation 4 also affects K4. K4 in turn must be squared (since the K expression takes the coefficient of the chemical equation and turns it into an exponent) 

So now (1/K¬3)(K4)2=overall K in my case. 

NaO2(s) + 2Na2O(s) ßà2NaO(g) + Na2O2(s) + Na(l) 
6.01×10-26 
Same concept here. For this one, its –reaction 1 + reaction 2 + reaction 3 + -reaction 4 
so that means (1/K1)(K2)(K3)(1/K4)

best me!

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Feb 1st, 2015

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