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

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O_{2}(g) + 4Na(l) 2Na_{2}O(s) K_{1} = 2.50×10^{49}
O_{2}(g) + 2Na(l) 2NaO(g) K_{2} = 2.50×10^{9}^{}
O_{2}(g) + 2Na(l) Na_{2}O_{2}(s) K_{3} = 2.00×10^{28}^{}
O_{2}(g) + Na(l) NaO_{2}(s) K_{4} = 3.33×10^{13}^{}
Wht would,be value equibrium constat each reactions,at 427°C?
Na_{2}O_{2}(s) +Na(l) NaO(g)+Na_{2}O(s)
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)
5.54×102
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)
6.01×1026
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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