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
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.