Respuesta :
Answer:
It will take 1.59 seconds of time to the concentration of [tex]H_3PO_4[/tex] to decrease to 20.0% of its initial value.
Explanation:
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
Half life for second order kinetics is given by:
[tex]t_{\frac{1}{2}=\frac{1}{k\times a_0}[/tex]
[tex]t_{\frac{1}{2}[/tex] = half life
k = rate constant
[tex]a_0[/tex] = initial concentration =
[tex]15min=\frac{1}{k\times 100}[/tex]
[tex]k=\frac{1}{1500}[/tex]
a= concentration left after time t
We have :
[tex]2H_3PO_4 (aq)\rightarrow P_2O_5 (aq) + 3H_2O (aq)[/tex]
Rate of the reaction, : [tex]R = (46.6 M^{-1}s^{-1})\times [H_3PO_4]^2[/tex]
Order of the reaction = 2
k = [tex]46.6 M^{-1} s^{-1}[/tex]
[tex][a_o]=0.660 M[/tex]
[tex][a]=20\% \times [a_o]=0.02\times 0.660 M=0.0132 M[/tex]
t = ?
[tex]\frac{1}{0.0132 M}=46.6 M^{-1} s^{-1}t+\frac{1}{0.660 M}[/tex]
t = 1.59 seconds
It will take 1.59 seconds of time to the concentration of [tex]H_3PO_4[/tex] to decrease to 20.0% of its initial value.
