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:

$t_{1/2}=\frac{[A_o]}{2k}$

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:

$t_{1/2}=\frac{5.90}{2\times 0.725}$

$t_{1/2}=4.069min=244.14s\approx 244s$

conversion used : (1 min = 60 s)

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