Respuesta :
Answer:
A) 0.065 M is its molarity after a reaction time of 19.0 hour.
B) In 52 hours [tex][Co(NH_3)5Br]^{2+}[/tex] will react 69% of its initial concentration.
Explanation:
[tex]Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)[/tex]
The reaction is first order in [tex][Co(NH_3)5Br]^{2+}[/tex]:
Initial concentration of [tex][Co(NH_3)5Br]^{2+}[/tex]= [tex][A_o]=0.100 M[/tex]
a) Final concentration of [tex][Co(NH_3)5Br]^{2+}[/tex] after 19.0 hours= [tex][A][/tex]
t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)
Rate constant of the reaction = k = [tex]6.3\times 10^{-6} s^{-1}[/tex]
The integrated law of first order kinetic is given as:
[tex][A]=[A_o]\times e^{-kt}[/tex]
[tex][A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}[/tex]
[tex][A]=0.065 M[/tex]
0.065 M is its molarity after a reaction time of 19.0 h.
b)
Initial concentration of [tex][Co(NH_3)5Br]^{2+}[/tex]= [tex][A_o]=x[/tex]
Final concentration of [tex][Co(NH_3)5Br]^{2+}[/tex] after t = [tex][A]=(100\%-69\%) x=31\%x=0.31x[/tex]
Rate constant of the reaction = k = [tex]6.3\times 10^{-6} s^{-1}[/tex]
The integrated law of first order kinetic is given as:
[tex][A]=[A_o]\times e^{-kt}[/tex]
[tex]0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}[/tex]
t = 185,902.06 s = [tex]\frac{185,902.06 }{3600} hour = 51.64 hours[/tex] ≈ 52 hours
In 52 hours [tex][Co(NH_3)5Br]^{2+}[/tex] will react 69% of its initial concentration.
