The rate of decomposition of XO2 after 1 hour : [tex]\tt A=A_o.e^{-1.08}[/tex]
Further explanation
Given
the half-life 38.6 min
time of decomposition = 1 hour
Required
the rate of decomposition
Solution
First-order reaction :
[tex]\tt A=A_o.e^{-kt}[/tex]
the half life=t1/2 :
[tex]\tt t\frac{1}{2}=\dfrac{ln~2}{k}[/tex]
so the rate constant (k) :
[tex]\tt k=\dfrac{ln~2}{38.6}=0.018[/tex]
The rate after 1 hour=60 min :
[tex]\tt A=A_o.e^{-0.018\times 60}\\\\A=A_o.e^{-1.08}[/tex]
