Question

acetic acid and water react to form hydronium cation and acetate anion, like this:

hch₃co₂(aq)+h₂o(l)→h₃o⁺(aq)+ch₃co₂⁻(aq)

imagine 249. mmol of ch₃co₂⁻ are removed from a flask containing a mixture of hch₃co₂, h₂o, h₃o⁺ and ch₃co₂⁻ at equilibrium, and then answer the following questions.

what is the rate of the reverse reaction before any ch₃co₂⁻ has been removed from the flask?

zero.

greater than zero, but less than the rate of the forward reaction.

greater than zero, and equal to the rate of the forward reaction.

greater than zero, and greater than the rate of the forward reaction.

what is the rate of the reverse reaction just after the ch₃co₂⁻ has been removed from the flask?

zero.

greater than zero, but less than the rate of the forward reaction.

greater than zero, and equal to the rate of the forward reaction.

greater than zero, and greater than the rate of the forward reaction.

what is the rate of the reverse reaction when the system has again reached equilibrium?

zero.

greater than zero, but less than the rate of the forward reaction.

greater than zero, and equal to the rate of the forward reaction.

greater than zero, and greater than the rate of the forward reaction.

how much less ch₃co₂⁻ is in the flask when the system has again reached equilibrium?

none.

some, but less than 249. mmol.

1. mmol.

more than 249. mmol.

Explanation:

Step1: Analyze the system before removal

At equilibrium, the rates of the forward and reverse reactions are equal. So, the rate of the reverse reaction before any \(CH_3CO_2^-\) has been removed from the flask is greater than zero, and equal to the rate of the forward reaction.

Step2: Analyze the system just after removal

When \(CH_3CO_2^-\) is removed, the reverse reaction (which consumes \(H_3O^+\) and \(CH_3CO_2^-\) to form \(HCH_3CO_2\) and \(H_2O\)) slows down. The forward reaction (which forms \(H_3O^+\) and \(CH_3CO_2^-\)) is still going at the rate it was at equilibrium before removal. So the rate of the reverse reaction just after the \(CH_3CO_2^-\) has been removed from the flask is greater than zero, but less than the rate of the forward reaction.

Step3: Analyze the system at new equilibrium

At equilibrium (including the new equilibrium), the rates of the forward and reverse reactions are equal. So the rate of the reverse reaction when the system has again reached equilibrium is greater than zero, and equal to the rate of the forward reaction.

Step4: Analyze the amount of \(CH_3CO_2^-\) at new equilibrium

According to Le - Chatelier's principle, when \(CH_3CO_2^-\) is removed, the system will shift to the right (forward reaction) to produce more \(CH_3CO_2^-\). But it can't completely replenish the \(249\) mmol that was removed. So there is some \(CH_3CO_2^-\) less than \(249\) mmol.

Answer:

1. Greater than zero, and equal to the rate of the forward reaction.
2. Greater than zero, but less than the rate of the forward reaction.
3. Greater than zero, and equal to the rate of the forward reaction.
4. Some, but less than 249 mmol.