Question

Suppose the surface-catalyzed hydrogenation reaction of an unsaturated hydrocarbon has a rate constant of 0.725 M/min. The reaction is observed to follow zero-order kinetics. If the initial concentration of the hydrocarbon is 5.90 M, what is the half-life of the reaction in seconds? *Please report 3 significant figures. Numbers only, no unit. No scientific notation.

Answer 1

Answer : The half-life of the reaction in seconds is, 244

Explanation :

The expression used for zero order reaction is:

`t_(1/2)=([A_o])/(2k)`

where,

`t_(1/2)` = half-life of the reaction = ?

`[A_o]` = initial concentration = 5.90 M

k = rate constant = 0.725 M/min

Now put all the given values in the above formula, we get:

`t_(1/2)=(5.90)/(2* 0.725)`

`t_(1/2)=4.069min=244.14s\approx 244s`

conversion used : (1 min = 60 s)

Thus, the half-life of the reaction in seconds is, 244