Solveeit Logo

Question

Question: In a hydrolysis reaction, \( 5{\text{g}} \) ethyl acetate is hydrolysed in presence of dilute \( HCl...

In a hydrolysis reaction, 5g5{\text{g}} ethyl acetate is hydrolysed in presence of dilute HClHCl in 300 minutes. If the reaction is of first order and the initial concentration of ethyl acetate is 22 gL122{\text{ g}}{{\text{L}}^{ - 1}} . Calculate the rate constant of the reaction.

Explanation

Solution

The rate constant, given by KK quantifies the speed of a chemical reaction with the assumption that the reaction is taking place throughout the volume of the solution involved in the chemical reaction. We shall substitute the given values in the formula below.

Formulas used: We will be using the formula for rate constant of first order reaction, which is given by, K=2.303tlog(aax)K = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)
where KK is the rate constant of the reaction, tt is the time taken for the reaction to occur, aa is the initial concentration of the reactant during the reaction, axa - x is the final concentration of the reactant after the reaction occurs.

Complete Step by Step answer
We know that when a chemical reaction takes place, the reactant is converted into the product over a short amount of time. The speed at which the conversion of the reactant into the product takes place is calculated by the rate constant of the reaction.
Here we can see that the above reaction, which is the hydrolysis of ethyl acetate in the presence of dilute HClHCl in 300 minutes time, is a reaction of the first order. A first order reaction is nothing but a reaction that depends only on the concentration of one of the reactants, there are either no other reactants or the existing ones do not affect the rate of the reaction.
So, we know that the reaction’s rate here solely depends on the concentration of ethyl acetate. We also know that the initial concentration of ethyl acetate in the reaction is a=22 gL1a = 22{\text{ g}}{{\text{L}}^{ - 1}} , we also know that in time t=300mint = 300\min 5g5{\text{g}} of ethyl acetate is hydrolysed. Thus, the concentration of ethyl acetate after 300 minutes will be, ax=225=17 gL1a - x = 22 - 5 = 17{\text{ g}}{{\text{L}}^{ - 1}} .
Substituting the value, we know in the equation to find the reaction rate,
K=2.303300log(2217)K = \dfrac{{2.303}}{{300}}\log \left( {\dfrac{{22}}{{17}}} \right)
Solving for KK we get,
K=2.303300×0.112K = \dfrac{{2.303}}{{300}} \times 0.112
K=0.00767×0.112K = 0.00767 \times 0.112
K=0.0008597min1=8.5×104min1\Rightarrow K = 0.0008597{\min ^{ - 1}} = 8.5 \times {10^{ - 4}}{\min ^{ - 1}}
Thus, the rate constant of the reaction K=8.5×104min1K = 8.5 \times {10^{ - 4}}{\min ^{ - 1}} .

Note
The rate of reaction of first order reaction can only be found considering one of the reactants, if the reaction had been of the second order then, the rate of the reaction will be given by, r=K[A]2r = K{[A]^2} or r=K[A]x[B]yr = K{\left[ A \right]^x}{\left[ B \right]^y} where the sum of xx and yy is always 2, since the reaction is of second order.