Question

The rate law for the decomposition of aqueous hydrogenperoxide

(H2O2) at 70 degrees celcius is first order in
with k=.0347 min to the negative 1.
A) Calculate the half-life (t1/2) for this reaction at
70degrees C
B) Given that the initial concentration of H202 is
.300Mcalculate the concentration of H2O2 after 60 minutes
haselapsed.

Answer 1

`t_(1/2)=19.98\ min`

`[A_t]=0.037\ M`

Step-by-step explanation:

(a)

Given that:

Rate constant, k = 0.0347 min⁻¹

The expression for the half-life is:-

`t_(1/2)=(\ln 2)/(k)`

Where, k is rate constant

So,

`t_(1/2)=(\ln 2)/(0.0347)\ min`

`t_(1/2)=19.98\ min`

(b)

Using integrated rate law for first order kinetics as:

`[A_t]=[A_0]e^(-kt)`

Where,

`[A_t]` is the concentration at time t

`[A_0]` is the initial concentration

Given that:

The rate constant, k =
`0.0347` min⁻¹

t = 60 min

Initial concentration
`[A_0]` = 0.300 M

Final concentration
`[A_t]` =?

Applying in the above equation, we get that:-

`[A_t]=0.300* e^(-0.0347* 60)\ M`

`[A_t]=0.3* (1)/(e^(2.082))\ M`

`[A_t]=0.037\ M`