SME
Analytical Chemistry
Kinetic methods
Use this for “With k = 0.250 M⁻¹s⁻¹ and the reagent at 0.400 M, 3.16 × 10⁻⁵ M of product forms in 12 s. Find the analyte’s starting concentration.”
Chart
[P](10) ≈ 3.16 × 10⁻⁵
This is an example. Type your own values to start a new problem.
k
Rate constant
example · M⁻¹s⁻¹
[R]
Reagent concentration
example · M
k′
Pseudo-first-order rate constant
calculated · s⁻¹
t
Fixed time
example · s
[P]_t
Product at time t
example · M
[A]₀
Starting analyte concentration
calculated · M
Formulas
k
′
=
k
×
[R]
0.1 = 0.25 × 0.4
[P]_t
=
[A]₀
×
(1
−
e
−
k
′ ×
t
)
3.16 ×
10
-5
=
5.00 ×
10
-5
×
(1
−
e
−0.1 × 10
)
Assumptions
•
The reagent R is in large excess, so [R] stays constant and the rate k[A][R] becomes pseudo-first order: k′ = k[R].
•
Each analyte particle gives one product particle, so [P]_t = [A]₀(1 − e−k′t).
•
Fixed-time method: every sample and standard is measured at the same time t, so [P]_t is proportional to [A]₀.
Step-by-step
WE KNOW
k = 0.25 M⁻¹s⁻¹, [R] = 0.4 M, t = 10 s, [P]_t = 3.16 × 10⁻⁵ M
FIND
pseudo-first-order rate constant (k′), starting analyte concentration ([A]₀)
Step 1 · Find pseudo-first-order rate constant (k′)
Use k′ = k × [R]
The reagent hardly changes, so fold its concentration into the rate constant.
k
′
=
0.25
×
0.4
k
′
=
0.1
s⁻¹
Step 2 · Find starting analyte concentration ([A]₀)
Use [P]_t = [A]₀ × (1 − e−k′ × t)
Divide
the
product
by
the
share
that
has
reacted
by
time
t
.
[A]₀
=
[P]_t
÷
(1
−
e
−
k
′ ×
t
)
[A]₀
=
3.16 ×
10
-5
1 −
e
−0.1 × 10
[A]₀
=
3.16 ×
10
-5
1 −
e
-1
[A]₀
=
3.16 ×
10
-5
1 − 0.3679
[A]₀
=
3.16 ×
10
-5
0.6321
[A]₀
=
5.00 ×
10
-5
M
→
k
′
t
=
1:
63.2%
of
the
analyte
has
reacted
by
the
fixed
time.
Check
Put the numbers back in. Both sides should match.
0.1 = 0.25 × 0.4 ✓
3.16 ×
10
-5
=
5.00 ×
10
-5
×
(1
−
e
−0.1 × 10
)
✓
← Previous lesson
Detection and quantitation limits
Next lesson →
Enzyme activity from an absorbance slope
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.