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Scott L.
Questions (3)
At a particular temp, K = 1.6 x 10^-5 for the reaction:
2SO3(g) <-> 2SO2(g) + O2 (g) If 4 mol of SO2 and 2 mol of O2 are placed
2 answers
744 views
Consider the reaction below:
4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain temp initially, all the reactants and products
2 answers
704 views
Consider the reaction below:
4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain temp initially, all the reactants and products
2 answers
777 views
Answers (3)
I don't understand where you got the 4 on the left side of your equation from? May you please explain?
Oh ok I understand now. Thanks!
i set up an ICE table then got: 10^80 = [(12+2x)^2 (12+6x)^6]/[(12-4x)^4 (12-3x)^3] But then it looks waaay to complicated. Am I doing it wrong? Please help. Thanks