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Question:

For the hydrolysis of methyl acetate in aqueous solution the above tabulated results were obtained: (a) Show that it follows a pseudo first-order reaction, as the concentration of water remains constant. (b) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given: log 2 = 0.3010, log 4 = 0.6021)

Solution:

(a) The reaction is 1st order w.r.t ester since [H₂O] remain constant. T₀ and Tt are concentrations at respective times. Thus, rate constant (k) = (1/t) ln(T₀ - Tt/T₀) ∴ k = (1/10) ln(0.10 - 0.05/0.10) = 0.0693 k = (1/10) ln(0.05 - 0.025/0.05) = 0.0693 s⁻¹ It appears from the calculation that the value of (1/t) ln(T₀ - Tt/T₀) is constant. Thus, the reaction is 1st order (as assumed) (b) The average rate constant is = Δ[CH₃COOCH₃]/Δt = [0.02 molL⁻¹ - 0.05 molL⁻¹]/20s - 10s = 0.025/10s molL⁻¹ = 0.0025 molL⁻¹s