Equilibrium Position: How Water Affects System

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Adding water to the system will increase the volume, leading to a decrease in concentration of the reactants and products. This dilution causes the reaction quotient Q to drop below the equilibrium constant K, initially favoring the forward reaction. However, as the solution becomes more diluted, the degree of dissociation of the complex increases, which can shift the reaction backward. The rates of the forward and backward reactions are affected differently by dilution, with the forward reaction rate decreasing more significantly. Overall, the equilibrium position is influenced by the dilution effect on concentration and dissociation.
nicolauslamsiu
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I want to ask how will the equilibrium position of the system change if water is added to the system?
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Co^2+(aq.) + 4cl^-1(aq.) =cocl4^2-(aq.)
 
The concentration will decrease because the volume will increase. This will make Q lower than the K value so the forward reaction should be favored(shifts right)
 
Will the value of Q become smaller the K if the concentration is lower?
 
*become greater
 
Can you rephrase your question, I'm not sure I understand.
 
AlphaEmission said:
The concentration will decrease because the volume will increase. This will make Q lower than the K value so the forward reaction should be favored(shifts right)

Quite the opposite, the more diluted the solution, the more dissociated the complex is.
 
By adding water the rate of the forward reaction decreases X to the power 2 times while the rate of the backward reaction will decrease only X times so the reaction moves backwards
 
Pretend the Keq is 1, and all species' molarity is 1 at the equilibrium. Now do the dilution recalculate Q based on the new concentrations. Compare Q with Keq and reason it out.
 
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And according to Ostwald's law : degree of dissociation increases by increasing dilution
 
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