The hydrolysis of ethyl acetate (CH3COOC2H5) by sodium hydroxide yields ethanol (C2H5OH) and sodium acetate (NaO2CCH3) by the reaction shown, which follows second order kinetics. In an experiment 500 mL of 0.1 M ethylacetate in water is mixed with 500 mL of 0.1 M NaOH in water. After 20 minutes the concentration of sodium acetate is measured and is 0.0435M. Determine the rate constant for the reaction.
(1/A)-(1/Ao) = kt
You know Ao and t from the problem. You can calculate A, subtitute and solve for k.
post your work if you get stuck.
To determine the rate constant for the reaction, we can use the integrated rate law for a second-order reaction. The integrated rate law equation is:
1/[A]t - 1/[A]0 = kt
Where:
[A]t = concentration of reactant at time t
[A]0 = initial concentration of reactant
k = rate constant
t = time
In this case, we are given the concentration of sodium acetate ([A]t) as 0.0435 M, the initial concentration of ethyl acetate ([A]0) as 0.1 M, and the time (t) as 20 minutes.
Let's plug the values into the integrated rate law equation and solve for k:
1/0.0435 - 1/0.1 = k * 20
Simplifying the equation:
23.0 - 10.0 = k * 20
13.0 = k * 20
Now, we can solve for k:
k = 13.0 / 20
k = 0.65 min^-1
Therefore, the rate constant for the reaction is 0.65 min^-1.
To determine the rate constant for the hydrolysis reaction, you can use the second-order rate equation:
Rate = k * [ethyl acetate] * [sodium hydroxide]
where:
- Rate is the rate of reaction
- k is the rate constant
- [ethyl acetate] and [sodium hydroxide] are the concentrations of ethyl acetate and sodium hydroxide, respectively.
Given that the reaction follows second-order kinetics, the rate equation can be simplified to:
Rate = k * [ethyl acetate]^2
To find the rate constant (k), we need to rearrange the equation and solve for k.
First, we need to determine the initial concentration of ethyl acetate ([ethyl acetate]_0). Since we mixed equal volumes of 0.1 M ethyl acetate and 0.1 M sodium hydroxide, the initial concentration of ethyl acetate is:
[ethyl acetate]_0 = (0.1 M) * (500 mL / 1000 mL)
= 0.05 M
Next, we can calculate the rate of the reaction using the concentration of sodium acetate after 20 minutes ([sodium acetate]_20 min) and the known volume:
Rate = (0.0435 M) / (20 min)
Finally, we can substitute the values into the rate equation to solve for the rate constant (k):
k = Rate / [ethyl acetate]^2
= (0.0435 M) / ((0.05 M)^2)
Evaluating this expression will give you the numeric value of the rate constant.