If Keq forward vs. reverse

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SaintJude

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If the equilibrium constant for a forward formation reaction is 6 x 10^5. What is the equilibrium of the decomposition (the reverse) reaction? 1/(6 x 10^5) ?
 
If the equilibrium constant for a forward formation reaction is 6 x 10^5. What is the equilibrium of the decomposition (the reverse) reaction? 1/(6 x 10^5) ?


Yup. Keq = [product]/[reactant] So if 10 = [product]/[reactant] then 0.1 = [reactant]/[product]
 
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