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If you added two equilibrium concentrations, the equilibrium constants would also be added together. However, most of the time when you modify existing equilibrium equations to find a new K value, you will multiply or divide the equilibrium equations because this allows you to remove terms from the numerator and denominator. I can't think of a reason to add equilibrium equations, but if you did, algebra dictates that you would add the equilibrium constants.
As stated in the above post, when you add together two equilibrium equations, the constants will also be added. However, often when we add two equilibrium equations together, we make slight modifications to get the end reaction equation we want. For example, to ensure the end result is balanced we might need to multiply the coefficients of one reaction by a number. When this occurs, we put the equilibrium constant of this equation to the power of that multiple. Then we would add this modified equilibrium constant to the constant of the other equation. Another example might entail adding the reverse reaction of one equation to the forward reaction of another. For the reverse reaction, we would have to modify the equilibrium constant of that equation by taking the inverse of it; then we would dadd this to the constant of the other equation.
They are added together. The examples we had in sapling added together 1/K or K^2 to show how if you reverse the reaction or double its moles then it still be added all together, but in a trickier way! Hope this helps.
Yes. When you increase the moles in an equilibrium equation, the moles of every reactant/product increase as well. So if you double the moles of one participant, you also have to double the moles of every other participant.
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