In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow

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+Pinkishorange(aq)+Br(aq)
The reaction is first order in Co(NH3)5Br2+, the rate constant at 25 C is 6.3×106 s1, and the initial concentration of Co(NH3)5Br2+ is 0.100 M.

What is its molarity after a reaction time of 37.0 h ?

How many hours are required for 79 % of the Co(NH3)5Br2+ to react?

Explanation / Answer

rate constant k = 6.3×10^6 s1

initial concentration Ao = 0.100 M

time = 37 hr = 133200 sec

k = 1/t ln (Ao / At)

6.3×10^-6 = 1/ 133200 ln (0.100 / At)

At = 0.0432

Molarity = 0.0432 M

2)

k = 1/t ln (Ao / At)

6.3×10^-6 = 1/ t ln (100 / 21)

t = 2.477 x 10^5 sec

time taken = 68.8 hrs

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