SME
General Chemistry II
Kinetics
Use this for “A first-order reaction has k = 5.00 × 10⁻⁴ s⁻¹. Starting at 0.200 M, find the concentration after 1000 s and the half-life.”
Diagram
Order 1: ln [A] = ln [A]₀ − kt, with k = 5 × 10⁻⁴ s⁻¹ and [A]₀ = 0.2 M.
t
1
2
=
ln 2
k
=
1,390
s,
the
same
each
time
(first
order).
At t = 1,000 s, [A] = 0.121 M.
The ln [A] plot is straight with slope −k: the test for order 1.
This is an example. Type your own values to start a new problem.
ln(
[A]₀ (M)
÷
[A] (M)
)
=
k (s⁻¹)
×
t (s)
t½
Half-life
calculated · s
f
Fraction left
calculated
Formulas
ln([A]₀
÷
[A])
=
k
×
t
ln(
0.2
0.121
)
=
5 ×
10
-4
×
1,000
t
1
2
=
ln(2)
k
1,390
=
ln(2)
5 ×
10
-4
f
=
[A]
÷
[A]₀
0.607
=
0.121
0.2
Assumptions
•
The reaction is first order in A: rate = k[A]. So [A] falls by the same fraction in every equal stretch of time.
•
The half-life t½ = ln 2 ÷ k does not depend on [A]₀: each half-life halves whatever is left.
•
A straight line of ln[A] against t, with slope −k, is the test for first order.
•
k and t use the same time unit, seconds here.
Step-by-step
WE KNOW
k = 5 × 10⁻⁴ s⁻¹, [A]₀ = 0.2 M, t = 1,000 s
FIND
concentration at time t ([A]), half-life (t½), fraction left (f)
Step 1 · Find concentration at time t ([A])
Use ln([A]₀ ÷ [A]) = k × t
Undo the ln: [A] is [A]₀ times e raised to −kt.
[A]
=
[A]₀
×
e
−
k
×
t
[A]
=
0.2
×
e
−(5 ×
10
-4
) × 1,000
[A]
=
0.2
×
e
−0.5
[A] = 0.2 × 0.6065
[A] = 0.1213 M
Step 2 · Find half-life (t½)
Use t½ = ln(2) ÷ k
At
t
1
2
,
[A]₀
÷
[A]
=
2,
so
kt
1
2
=
ln
2.
Divide
ln
2
by
k
.
t
1
2
=
ln(2)
5 ×
10
-4
t
1
2
=
0.6931
5 ×
10
-4
t
1
2
=
1,390
s
Step 3 · Find fraction left (f)
Use f = [A] ÷ [A]₀
Divide what is left by what there was at the start.
f
=
0.1213
0.2
f
=
0.607
Check
Put the numbers back in. Both sides should match.
ln(
0.2
0.1213
)
=
5 ×
10
-4
×
1,000
✓
1,390
=
ln(2)
5 ×
10
-4
✓
0.607
=
0.1213
0.2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
A zero-order reaction
Use this for “A zero-order reaction has k = 2.00 × 10⁻³ M/s. Starting at 0.500 M, find the concentration after 100 s, the half-life and when A runs out.”
Order and rate constant from initial rates
Use this for “With [B] held at 0.100 M, raising [A] from 0.100 M to 0.200 M raises the initial rate from 2.00 × 10⁻³ to 8.00 × 10⁻³ M/s. The reaction is first order in B. Find the order in A and k.”
Activation energy from two temperatures
Use this for “A rate constant doubles, from 1.00 × 10⁻³ s⁻¹ at 25 °C to 2.00 × 10⁻³ s⁻¹ at 35 °C. Find the activation energy and the frequency factor A.”
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Use this for “The slow step is NO₂ + NO₂ → NO₃ + NO, then NO₃ + CO → NO₂ + CO₂ is fast. What rate law fits?”