Le Chatelier's Principle

1447 WordsJul 15, 20186 Pages
Discussion: At Q 1 A reaction is at equilibrium when the ratio of the product and the reactant side are equal. This could further be seen by the reaction quotient depicted by Qc is equal to the equilibrium constant, Kc, of the given chemical reaction. Given that aA + bB ↔ cC +dD, the small letters being the mole coefficients and the big letters being the molecules of the chemical equation, K_eq=([〖C]〗^c [〖D]〗^d)/([〖A]〗^a [〖B]〗^b ) is the formula for finding the Kc of any given reversible equation. The symbol used in writing an equation that involves equilibrium is a double-headed arrow to indicate that indeed, the reaction could go forward and backward. At Q 2 Le Chatelier’s Principle states that a solution at equilibrium will shift…show more content…
Any shift in the equilibrium would be reverted back after the second solution would be added since the acid would just neutralize the base (and the base neutralize the acid.) When 1 M HCl was added to the 1 mL of 0.1 M K2CrO4, it caused the originally clear yellow solution into a yellow orange color. When 1 M NaOH was then added after, the solution returned to its original color which was clear yellow. The reverse occurred when 1 M NaOH was added to the 1 mL of 0.1 M K2Cr2O7 then 1 M HCl. When added with 1 M NaOH first, the originally clear orange solution became yellow orange then back to orange again when 1 M HCl was added. This shows that what ever acid or base was added, the counterpart was able to neutralize the added solution and revert the reaction. Whenever HCl is added to the solution, the HCl will dissociate into ions, thus increasing the supply of H+ that would cause a shift in the equilibrium. Similarly, this is what also occurs when NaOH is added to the solution, the NaOH would dissociate and the OH would then neutralize the H+ thus forming an excess of H2O present in the solution that would ultimately lead to a shift in the equilibrium. Based on the net equation CrO42-(aq) + H+(aq) ↔ Cr2O72-(aq) + H2O(l), when HCl is added to K2CrO4 the solution shifted right to form some K2Cr2O7 to relieve the stress from the excess H+ from the HCl added. This can be seen by the change
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