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Firstly, let me confirm that my definition for right shift is correct: a right shift entails an increased ratio of product to reactant.

Say you have the reaction A -> B. The standard notion is that increasing concentration of the reactant shifts the equation to the right but that can obviously not be the case here. Explanation: Take two systems composed of A -> B respectively and add them together. You'll end up with 2A -> 2B with the ratio remaining the same.

However, if you have the equation A + B -> C, what will happen is that a greater percentage of B will react, and then the equation will shift to the right relative to B, but to the left relative to A, and to the left relative to the equation as a whole. However, if you increase both A and B by the same factor no shift will occur.

My overarching question is then whether I'm correct stating the above, and whether as I've described shifts left and right occur not relative to a side of the equation but rather relative to a specific part of it.

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closed as unclear what you're asking by Mithoron, M.A.R. ಠ_ಠ, Todd Minehardt, airhuff, bon Feb 13 '18 at 13:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Your third paragraph is a bit unclear is " A + B -> C" in continuation with the earlier given paragraph, or are you considering a new traction altogether. Your arrows are not the ones for equilibrium. $\endgroup$ – Avnish Kabaj Feb 10 '18 at 14:09
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Your understanding of Le-Chatlier's principle is wrong. Whenever there is an increase in the concentration of reactants (Either A or B in your case), equilibrium shifts towards products. And remember one thing Le-Chatlier's principle states that whenever a change has occurred in equilibrium, it tries to oppose the change. So here when you increase the concentration of reactant either A or B or both, the equilibrium shifts towards products.

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Say you have the reaction A -> B. Standard notion is that increasing concentration of reactant shifts equation to the right but that can obviously not be the case here. Explanation: Take two systems composed of A -> B respectively and add them together. You'll end up with 2A -> 2B with the ratio remaining the same.

However, if you have the equation A + B -> C what will happen is that a greater percentage of B will be reacted and then equation will shift to the right relative to B but left relative to A. And left relative to the equation as a whole. Correct? However, if you increase both A and B by the same factor no shift will occur.

None of these statements are correct. In the reaction $\ce{A \to B}$, increasing the concentration of $\ce{A}$ does shift the equilibrium to the right. Mathematically,you look at this through the equilibrium constant, $$K = \frac{[B]}{[A]}$$ Adding more $\ce{A}$ has to be followed by an increase in $\ce[B]$ to preserve the equilibrium constant constant. The logic of relating the species concentrations to their stoichiometric coefficients is flawed - and I am also kind of confused at what you mean.

As to the paragraph after that, there is no such thing as a reaction shifting relative to a certain reactant. The reaction either shifts towards products or towards reactants, there is no relativity.

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The concentration of reactant A is not determined by its coefficient in an equation. The concentration of A is expressed in moles per liter and applies equally to situations where A-->B or 2A-->C. You measure (or set) the concentrations at the beginning of the experiment.

The equilibrium constant is different for each of those reactions, and indicates whether the effect of changing the concentration of A is first order (linear) or second order (concentration squared).

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