SME
General Chemistry II · Problem type
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.”
Diagram
Order 0: [A] = [A]₀ − kt, with k = 0.002 M/s and [A]₀ = 0.5 M.
t
1
2
=
[A]₀
÷
(2k)
=
125
s;
it
runs
out
at
[A]₀
÷
k
=
250
s.
At t = 100 s, [A] = 0.3 M.
[A] against t is itself the straight line, slope −k.
This is an example. Type your own values to start a new problem.
[A] (M)
=
[A]₀ (M)
−
k (M/s)
×
t (s)
t½
Half-life
calculated · s
tend
Time A runs out
calculated · s
Formulas
[A]
=
[A]₀
−
k
×
t
0.3 = 0.5 − 0.002 × 100
t
1
2
=
[A]₀
÷
(2
×
k
)
125
=
0.5
2 × 0.002
tend
=
[A]₀
÷
k
250
=
0.5
0.002
Assumptions
•
The reaction is zero order in A: rate = k, whatever [A] is, as when a catalyst surface is fully covered. So [A] falls by the same amount every second.
•
The half-life t½ = [A]₀ ÷ (2k) depends on [A]₀, and there is no second half-life: A runs out at tend = 2t½.
•
The law holds only until A runs out, so t is at most [A]₀ ÷ k. A straight line of [A] against t, with slope −k, is the test for zero order.
•
k, [A] and t use the same units: M and seconds here.
Step-by-step
WE KNOW
k = 0.002 M/s, [A]₀ = 0.5 M, t = 100 s
FIND
concentration at time t ([A]), half-life (t½), time A runs out (tend)
Step 1 · Find concentration at time t ([A])
Use [A] = [A]₀ − k × t
A
is
used
at
the
steady
rate
k
,
so
kt
is
used
in
time
t
.
Take
that
from
[A]₀.
[A] = 0.5 − 0.002 × 100
[A] = 0.5 − 0.2
[A] = 0.3 M
Step 2 · Find half-life (t½)
Use t½ = [A]₀ ÷ (2 × k)
Half
of
[A]₀
is
used
at
the
steady
rate
k
.
Divide
that
half
by
k
.
t
1
2
=
0.5
2 × 0.002
t
1
2
=
0.5
0.004
t
1
2
=
125
s
Step 3 · Find time A runs out (tend)
Use tend = [A]₀ ÷ k
All
of
[A]₀
is
used
at
the
steady
rate
k
.
Divide
[A]₀
by
k
.
tend
=
0.5
0.002
tend = 250 s
Check
Put the numbers back in. Both sides should match.
0.3 = 0.5 − 0.002 × 100 ✓
125
=
0.5
2 × 0.002
✓
250
=
0.5
0.002
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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