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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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Feb 1st, 2015
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Feb 1st, 2015
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