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
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)