Suppose the surface-catalyzed hydrogenation reaction of an unsaturated hydrocarb

Question

Suppose the surface-catalyzed hydrogenation reaction of an unsaturated hydrocarbon has a rate conga of 0.720 M/min. The reaction is observed to follow zero-order kinetics. If the initial concentration of the hydrocarbon is 2.78 M, what is the half-life of the reaction? The hydrolysis of methyl acetate in water, CH3COOC2H5(aq)+ H2O( ) CH3COOH(aq) + C2H5OH(aq), is pseudo-first order in dilute aqueous solutions with a rate constant of 2.00x 10-3 min-What is the value of the second-order rate constant. where the concentration of water is no longer assumed to be 55.5 M?

Explanation / Answer

1. We know that for zero order reaction the half life can be calculated as:

t1/2 = [A0] / 2K

t1/2 = ??

[A] = 2.78

K = 0.720 M / min

t1/2 = 2.78 / 2 X 0.72 = 1.93minutes

2. The ester hydrolysis is pseudo first order reaction

so rate = K [ester]

if it [water ] is not assumed to be constant then

rate = K' [ester][water]

so K' = K/ [water] = 2 X 10^-2 / 55.5= 0.036 X 10^-2

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