Time for completion of 75% of a reaction is thrice the time for completion of the same reaction.Hence order of the reaction is
Answers
Answer:
X
=
2
Explanation:
As you know, the rate of a first-order reaction depends linearly on the concentration of a single reactant. The rate of a first-order reaction that takes the form
A
→
products
can thus be written as
rate
=
−
d
[
A
]
d
t
=
k
⋅
[
A
]
, where
k
- the rate constant for the reaction
Now, in Integral form (I won't go into the derivation here), the rate law for a first-order reaction is equal to
ln
(
[
A
]
[
A
0
]
)
=
−
k
⋅
t
, where
[
A
]
- the concentration of the reactant after the passing of time
t
[
A
0
]
- the initial concentration of the reactant
Now, the half-life of the reaction is the time needed for the concentration of the reactant to reach half of its initial value.
You can say that
t
=
t
1/2
⇒
[
A
]
=
1
2
⋅
[
A
0
]
Plug this into the equation for the rate law to get
ln
⎛
⎝
1
2
⋅
[
A
0
]
[
A
0
]
⎞
⎠
=
−
k
⋅
t
1/2
This means that you have
t
1/2
=
ln
(
1
2
)
−
k
=
ln
(
1
)
−
ln
(
2
)
−
k
=
ln
(
2
)
k
In your case, you want to figure out how many half0lives must pass in order for
75
%
of the reactant to be consumed. In other words, you need
25
%
of the reactant to remain after a time
t
x
t
=
t
x
⇒
[
A
]
=
25
100
⋅
[
A
0
]
=
1
4
⋅
[
A
0
]
Once again, plug this into the rate law equation to get
ln
⎛
⎝
1
4
⋅
[
A
0
]
[
A
0
]
⎞
⎠
=
−
k
⋅
t
x
This means that you have
t
x
=
ln
(
1
4
)
−
k
=
ln
(
1
)
−
ln
(
4
)
−
k
=
ln
(
4
)
k
But since
ln
(
4
)
=
ln
(
2
2
)
=
2
⋅
ln
(
2
)
you can say that
t
x
=
2
⋅
t
1/2
ln
(
2
)
k
=
2
⋅
t
1/2
Therefore,
X
=
2