Given:
[A]0 = 0.80 M, [A] = 0.50 M, t = 15 min, second-order reaction
Second-order: 1/[A] - 1/[A]0 = k * t
1/[A] - 1/[A]0 = 1/0.50 - 1/0.80 = 2 - 1.25 = 0.75 M ¹
k = (1/[A] - 1/[A]0)/t = 0.75 / 15 = 0.05 M ¹·min ¹

Given,

Initial concentration:
[
A
]
0
=
0.80

M
[A]
0

=0.80M
Final concentration:
[
A
]
=
0.50

M
[A]=0.50M
Time:
t
=
15

min
t=15min
Reaction type: Second-order
Step 1: Use the second-order formula
For a second-order reaction:
1
[
A
]

1
[
A
]
0
=
k
t
[A]
1


[A]
0


1

=kt
Step 2: Substitute the values
1
0.50

1
0.80
0.50
1


0.80
1


Calculate each term:
1
0.50
=
2
0.50
1

=2
1
0.80
=
1.25
0.80
1

=1.25
Now subtract:
2

1.25
=
0.75
2 1.25=0.75
So,
k
t
=
0.75
kt=0.75
Step 3: Solve for
k
k
k
=
0.75
15
k=
15
0.75


k
=
0.05
k=0.05
Final Answer:
k
=
0.05

M

1

m
i
n

1
k=0.05M
1
min
1


In Simple Terms:
The reaction’s rate constant is 0.05 per molarity per minute, meaning the reaction speed depends on both concentration and time in a second-order way.

1/[A]t - 1/[A]0 = kt.

Given: [A]0 = 0.80 M, [A]t = 0.50 M, and t = 15 minutes.

1/0.50 M - 1/0.80 M = k * 15 min.
2.0 M^-1 - 1.25 M^-1 = k * 15 min.
0.75 M^-1 = k * 15 min.
k = 0.75 M^-1 / 15 min = 0.05 M^-1 min^-1.

Therefore, the rate constant of the reaction is 0.05 M^-1 min^-1.