Question

The hydrolysis of ethylacetate by sodium hydroxide at 315 K, follows a second order kinetics with specific rate constant, K = 2.1 mol^-¹dm^³s-¹. If the initial concentration of ethylacetate is 0.04 moldm-³, Calculate the concentration of ethylacetate remaining after 85 seconds

Answer 1

The rate law for the hydrolysis reaction is given by:

Rate = k[ethylacetate]^2

We are given that the specific rate constant, k = 2.1 mol^-1dm^3s^-1.

The initial concentration of ethylacetate, [ethylacetate]₀ = 0.04 mol dm^-³.

We are asked to find the concentration of ethylacetate remaining after 85 seconds.

Let the concentration of ethylacetate remaining after 85 seconds be [ethylacetate]_t.

Using the integrated rate law for a second-order reaction:

1/[ethylacetate]_t = 1/[ethylacetate]₀ + kt

Plugging in the values:

1/[ethylacetate]_t = 1/0.04 + (2.1 mol^-1dm^3s^-1 × 85 s)

1/[ethylacetate]_t = 25 + 178.5

1/[ethylacetate]_t = 203.5

Therefore, [ethylacetate]_t = 1/203.5 = 0.0049 mol dm^-³.

The concentration of ethylacetate remaining after 85 seconds is 0.0049 mol dm^-³.