Question

Acetic acid and ethanol react to form ethyl acetate and water.If 94.0 mmol of C2H5CO2CH3 are removed from a flask containing a mixture of HCH3CO2, C2H5OH, C2H5CO2CH3 and H2O at equilibrium, then following questions are to be answered. 1. What is the rate of the reverse reaction before any C2H5CO2CH3 has been removed from the flask?2. What is the rate of the reverse reaction just after the C2H5CO2CH3 has been removed from the flask?3. What is the rate of the reverse reaction when the system has again achieved equilibrium?4. How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?

Answer 1

Step-by-step explanation:

The equation for the react between Acetic acid and ethanol to form ethyl acetate and water is :

`HCH_3CO_2_((aq))+C_2H_5OH_((aq)) \to C_2H_5CO_2CH_3_((aq)) + H_2O_((l))`

Imagine if 94.0 mmol of
`C_2H_5CO_2CH_3` are removed from a flask; Then:

We are to answer the following questions:

1. What is the rate of the reverse reaction before any
`C_2H_5CO_2CH_3` has been removed from the flask?

The reaction above is called an esterification reaction;

So the rate of reverse reaction before any
`C_2H_5CO_2CH_3` is removed is greater than zero and equal to forward reaction rate.

2. What is the rate of the reverse reaction just after the
`C_2H_5CO_2CH_3` has been removed from the flask?

Just after the
`C_2H_5CO_2CH_3` has been removed from the flask, the rate of the reverse reaction is greater than zero but less than forward reaction rate.

3. What is the rate of the reverse reaction when the system has again achieved equilibrium?

When the system has again achieved equilibrium, the rate of the reverse reaction is greater than zero and equal to forward reaction rate because we it has achieved the equilibrium, hence, the reaction tends to proceed in the forward direction.

4. How much less
`C_2H_5CO_2CH_3` is in the flask when the system has again reached equilibrium?

The
`C_2H_5CO_2CH_3` in the flask when the system has again reached equilibrium is lesser by 94.0 mmol as given right from the question