Question

In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced by a water molecule, yielding the pinkish-orange complex ion : Co(NH3)5(H2O)3+
Co(NH3)5Br2+Purple(aq)+H2O(l)? Co(NH3)5(H2O)3+Pinkish?orange(aq)+Br?(aq...
The reaction is first order in Co(NH3)5Br2+, the rate constant at 25 ?C is 6.3�10?6 s?1, and the initial concentration of Co(NH3)5Br2+ is 0.100 M.

A; What is its molarity after a reaction time of 19.0h ?

B; How many hours are required for 69% of the Co(NH3)5Br2+ to react?

Answer 1

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$:

Initial concentration of $[Co(NH_3)5Br]^{2+}$= $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of $[Co(NH_3)5Br]^{2+}$= $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.