Question

The rate law for the decomposition of aqueous hydrogenperoxide

Answer 1

Answer:

$t_{1/2}=19.98\ min$

$[A_t]=0.037\ M$

Explanation:

(a)

Given that:

Rate constant, k = 0.0347 min⁻¹

The expression for the half-life is:-

$t_{1/2}=\frac{\ln 2}{k}$

Where, k is rate constant

So,  

$t_{1/2}=\frac{\ln 2}{0.0347}\ min$

$t_{1/2}=19.98\ min$

(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:-

$[A_t]=0.300\times e^{-0.0347\times 60}\ M$

$[A_t]=0.3\times \frac{1}{e^{2.082}}\ M$

$[A_t]=0.037\ M$