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I know that if we add 1 mole to reactant side then the reaction will shift forward and if on product side the it will shift backward according to Le Chatelier's principle.Then what will happen if I add 1 mole on both sides?Please explain.

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    $\begingroup$ Specify exactly what are you adding to the reactants and products side. Do you mean you are adding 1 mole of PCl3 and Cl2 both on the products side $\endgroup$ – Daenys Targaryen May 3 '17 at 5:20
  • $\begingroup$ I mean that I know when we increase the concentration of reactants the reaction will do forward and if I increase concentration of products the it will go backward.Now where will it go if I add 1 mole of PCl5,1 mole of Cl2 and 1 mole of PCL3 in the equilibrium mixture $\endgroup$ – Pranav May 3 '17 at 6:25
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    $\begingroup$ Write Kc expression for the equilibrium, now write Qc expression when you add 1 mole of each of your reactants and products in the mixture. If Qc<Kc the equilibrium will move forward and vice versa $\endgroup$ – Daenys Targaryen May 3 '17 at 6:31
  • $\begingroup$ Can we tell it directly by Le Chatliers principle without knowing the concentrations of the mixture $\endgroup$ – Pranav May 3 '17 at 6:52
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    $\begingroup$ Le Chatelier's principle would have helped if you were changing the equilibrium concentration of either the reactants or products(it's very qualitative). To be sure you could always approach the problem In a standard way,that is what I just described. $\endgroup$ – Daenys Targaryen May 3 '17 at 8:31
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In a reaction of the form $\ce{A <=> B + C }$ if $a_0$ is the initial amount of A (and B and C are initially zero) then if an amount x reacts the equilibrium constant $K_e = x_e^2/(a_0-x_e) $ where $x_e$ is the equilibrium amount reacted so equilibrium A is $a_e = a_0-x_e$. Solving gives $x_e = (-K_e+\sqrt{K_e^2+4K_ea_0})/2$.

Now adding 1 mol to each side gives $K_e= (x_e'+1)^2/(a_0+1-x_e')$. Solving again gives $x_e' = (-K_e-2+\sqrt{K_e^2+4K_ea_0+8K_e})/2$. You can work out the ratio $x_e'/x_e$ for different values of $k_e$ for simplicity, say, with $a_0=1$.

You can repeat the calculation for different initial amounts of B and C and the equations are far more messy but the final result similar.

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