SME
General Chemistry II · Problem type
A second-order reaction
Use this for “A second-order reaction has k = 0.500 M⁻¹s⁻¹. Starting at 0.100 M, find the concentration after 60 s and the first half-life.”
Diagram
Order 2: 1/[A] = 1/[A]₀ + kt, with k = 0.5 M⁻¹s⁻¹ and [A]₀ = 0.1 M.
t
1
2
=
1
k[A]₀
=
20
s,
and
each
half-life
is
twice
the
last.
At t = 60 s, [A] = 0.025 M.
The 1/[A] plot is straight with slope +k: the test for order 2.
This is an example. Type your own values to start a new problem.
1
[A] (M)
=
1
[A]₀ (M)
+
k (M⁻¹s⁻¹)
×
t (s)
t½
First half-life
calculated · s
Formulas
1
÷
[A]
=
1
÷
[A]₀
+
k
×
t
1
0.025
=
1
0.1
+
0.5
×
60
t
1
2
=
1
k
× [A]₀
20
=
1
0.5 × 0.1
Assumptions
•
The reaction is second order in A: rate = k[A]². So it slows down much faster than a first-order reaction as A is used up.
•
The half-life t½ = 1 ÷ (k[A]₀) depends on [A]₀: each half-life is twice as long as the one before.
•
A straight line of 1/[A] against t, with slope +k, is the test for second order.
•
k, [A] and t use the same units: M and seconds here.
Step-by-step
WE KNOW
k = 0.5 M⁻¹s⁻¹, [A]₀ = 0.1 M, t = 60 s
FIND
concentration at time t ([A]), first half-life (t½)
Step 1 · Find concentration at time t ([A])
Use 1 ÷ [A] = 1 ÷ [A]₀ + k × t
Add kt to 1/[A]₀ to get 1/[A], then flip it over.
[A]
=
1
1 ÷ [A]₀ +
k
×
t
[A]
=
1
1 ÷ 0.1 + 0.5 × 60
[A]
=
1
10 + 30
[A]
=
1
40
[A] = 0.025 M
Step 2 · Find first half-life (t½)
Use t½ = 1 ÷ (k × [A]₀)
At
t
1
2
,
[A]
=
[A]₀
÷
2,
so
1/[A]
rises
by
1/[A]₀.
Divide
1
by
k
times
[A]₀.
t
1
2
=
1
0.5 × 0.1
t
1
2
=
1
0.05
t
1
2
=
20
s
Check
Put the numbers back in. Both sides should match.
1
0.025
=
1
0.1
+
0.5
×
60
✓
20
=
1
0.5 × 0.1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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