For a second order reaction, the half life is 726 seconds. Starting with a concentrate of 0.600 m, how long will it take for the concentration to decrease to 0.150m
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Answer:
Explanation:
We are given the following data;
Initial concentration = 0.600M
Final concentration = 0.150M
half=life = 726 seconds
So, let k be reaction constant,
t
1
/
2
to be half-life,
[
A
]
be the reactant concentration at time t and
[
A
]
∘
be the initial reaction concentration.
The second-order reaction will have a reaction constant k as follows;
k=
1
(
t
/
2
×
[
A
]
∘
)
=
1
(
726
×
0.600
)
=
0.002296
M
−
1
s
−
1
When
[
A
]
= 0.150M
Then;
1
[
A
]
=
1
[
A
]
∘
=
k
t
=>
1
0.150
=
1
0.600
+
0.002296
t
1
0.150
−
1
0.600
=
0.002296
t
t = 2178s
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