Question

The half-life of a reaction, t1/2, is the time required for one-half of a reactant to be consumed. It is the time during which the amount of reactant or its concentration decreases to one-half of its initial value.
Determine the half-life for the reaction in Part B using the integrated rate law, given that the initial concentration is 1.90mol?L?1 and the rate constant is 0.0016mol?

Answer 1

Answer: The half life of the reaction is 593.8 seconds

Explanation:

We are given:

Rate constant = $0.0016mol/L.s$

The formula for determining the unit of 'k' is:

$\text{Unit}=\frac{(Concentration)^{1-n}}{Time}$

where, n = order of reaction

The unit of concentration is, M or mole/L

The unit of time is, second or 's'

Evaluating the value of 'n' from above equation:

$mol.L^{-1}s^{-1}=\frac{(mol/L)^{1-n}}{s}\\\\n=0$

The reaction is zero order reaction.

The equation used to calculate half life for zero order kinetics:

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

where,

k = Rate constant = $0.0016mol/L.s$

$[A_o]$ = initial concentration = 1.90 mol/L

Putting values in above equation, we get:

$t_{1/2}=\frac{1.90mol/L}{2\times 0.0016mol/L.s}=593.8s$

Hence, the half life of the reaction is 593.8 seconds