Rate law for the reaction A + 2B → C is found to be Rate = k [A][B.] Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be______
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Answer:
you can see, the rate is proportional to the concentration of
A
and the square of the concentration of
B
.
Let's say that initially, you have
[
A
]
=
x
and
[
B
]
=
y
. The rate law in this case is equal to
rate 1
=
k
⋅
x
⋅
y
2
Now you double the concentrations of
A
and
B
to new values of
2
x
and
2
y
. The new rate law will be
rate 2
=
k
⋅
(
2
x
)
⋅
(
2
y
)
2
rate 2
=
k
⋅
2
x
⋅
4
y
2
=
8
⋅
x
⋅
y
2
As you can see, you have
rate 2
rate 1
=
k
⋅
8
⋅
x
⋅
y
2
k
⋅
x
⋅
y
2
=
8
Therefore, the rate of the reaction has increased by a factor of
8
.
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