R-ATCHE-5

Equilibrium for Concentrations

• When equilibrium is re-established, concentration will tend to be different.
  • There are funny exceptions of course :(

$2CrO_{4(aq)} ^{2-}+2H^{+}{(aq)}\rightleftharpoons Cr{2}O_{7(aq)}^{2-}+H_{2}O_{(l)}$

• If we increase the concentration of anything in this system, equilibrium will be disrupted and thus the system + other concentrations will change to re-establish equilibrium.

• Suppose we add concentrated $HCl$.

• What we have done is increased the concentration of a reagent, specifically $[H^+]$.

Initial change

• Rate of forward reaction increased
  • The increased concentration of reagents means more collisions occur between reagents hence more successful collisions occur between reagents, and the rate of the forward reaction thus is increased.
• Rate of reverse reaction unchanged

What happens next?

Logically, as the increased rate of forward reaction is initially faster than the reverse, when a new equilibrium is established, the forward rate will be faster than the reverse.

• It is important to say that the forward reaction is favoured.

As the forward rate is faster than the reverse rate, the system is consuming more reagents than it produces. Thus the concentration of reagents falls.

As the reverse rate is slower than the forward rate, the system consumes less products than it produces, hence the concentration of products rises.

As a result of the changing concentrations, the forward rate will partially decrease as the concentration of reagents fall, and the reverse rate will partially increase as the concentration of products rises.

This occurs until the rates are equal, and equilibrium is thus re-established having shifted right.

Things to consider

• Both reaction rates are faster than they were at the previous equilibrium.
• When doing questions, distinguish between what happens "at the time of change", and "when equilibrium is re-established".
  • At the time of change refers to the initial change. i.e. when we increase the concentration of a reagent, what is the initial change
    • Forward reaction rate is increased
    • Reverse reaction rate is unaffected
  • When equilibrium is re-established refers to after the initial change. Say the initial change was an increase in the concentration of the reagent.
    • Both reverse and forward reaction rates are higher than at the initial equilibrium
    • $[H^+]$ is lower than it was when more was added, but higher than it was at the initial equilibrium.
    • $[CrO_{4}^{2-}]$ is lower than at the initial equilibrium
    • $[Cr_{2}O_{7}^{2-}]$ is higher than at the initial equilibrium
      • Note: the difference in concentrations from initial to final is proportional to the molar ratio of the equation. This should be shown on your graph!

Review: Equilibrium Constant

• Note: $K_{c}$ only changes when temperature changes, otherwise it is constant!

Recall the equilibrium system $2CrO_{4(aq)}^{2-}+2H^{+}{(aq)}\rightleftharpoons Cr{2}O_{7(aq)}^{2-}+H_{2}O_{(l)}$

Hence, $K_{c} = \frac{[Cr_{2}O_{7}^{2-}]}{[CrO_{4}^{2-}]^2[H^{+}]^2}$

Le Chatelier's Principle

If we increase $[H^+]$ the forward reaction is favoured to oppose the change(consume $[H^+]$). This causes an increase in products, equilibrium position shifts right.

• $[H^+]$ is still greater than before - partially counteracted.

Decreasing Concentrations

Recall the equilibrium system $2CrO_{4(aq)}^{2-}+2H^{+}{(aq)}\rightleftharpoons Cr{2}O_{7(aq)}^{2-}+H_{2}O_{(l)}$

• Add $OH^-$ - This will reduce $[H^+]$.

Initial Change

• Concentration of reagents falls
• Concentration of products is unchanged

It follows that:

• Forward rate decreases
• Reverse rate is unchanged

Use Le Chatelier's Principle

• A decrease in $[H^+]$ causes the reverse reaction to be favoured (produce more $H^+$)
• This causes an increase in reactants, and equilibrium position shifts left.

Final Change

• Both rates of reaction are lower than at the initial equilibrium
• $[H^+]$ increases but not to its previous level
• $[CrO_{4}^{2-}]$ increases by the same amount as $[H^+]$
• $[Cr_{2}O_{7}^{2-}]$ is decreased by half the amount of $[H^+]$
  • Note the specific amount is based on the molar ratios.

Dilution

When we dilute an aqueous system, the concentration of all products are decreased.

Initial Change

• Forward and reverse rates are decreased.

Recall the equilibrium system $2CrO_{4(aq)}^{2-}+2H^{+}{(aq)}\rightleftharpoons Cr{2}O_{7(aq)}^{2-}+H_{2}O_{(l)}$

There must be a winner, cause of course there must.

Hence, one of the reactions will slow down more and one will slow down less. This is the favoured reaction(slows down less).

Le Chatelier's Principle

Overall ion concentration is decreased. The reverse reaction produces more moles compared to the forward reaction(4-1). Hence the reverse reaction is favoured so more ions are produced, and thus partially counteracting the decreased overall ion concentration.

Why does this happen?

• The forward reaction required the collision of 4 different ions

• The reverse reaction requires the collision of 2 different species.

• Hence, when the concentration of everything is decreased, the forward reaction is much more impacted by the dilution as it is even harder to have the 4 ions collide.

Important Note

• I will be referring to Le Chatelier's Principle as Le Shat from now on. Both because it is easier and it is funny.

• We usually say partially counteract, but we must say only counteract for special cases. i.e. decomposition of $CaCO_{3}$.

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