SME
Grade 10 Science · Problem type
Average reaction rate
Use this for “[A] falls from 1.20 mol/L at 30 s to 0.75 mol/L at 120 s. What is the average rate?”
Diagram
The curve (drawn first-order, for its shape) falls fastest at the start.
The secant through (30 s, 1.2 mol/L) and (120 s, 0.75 mol/L) has slope Δ[A]/Δt = −0.45/90 = −0.005.
The average rate is −Δ[A]/Δt = 0.005 mol/(L·s).
This is an example. Type your own values to start a new problem.
r (mol/(L·s))
=
−
[A]₂ (mol/L)
−
[A]₁ (mol/L)
t₂ (s)
−
t₁ (s)
Δt
Time between
calculated · s
Δ[A]
Change in concentration
calculated · mol/L
Formulas
Δt
=
t₂
−
t₁
90 = 120 − 30
Δ[A] = [A]₂ − [A]₁
−0.45 = 0.75 − 1.2
r
=
−Δ[A]/Δt
0.005 = −(−0.45)/90
Assumptions
•
A reactant is used up, so its concentration falls: the rate is −Δ[A]/Δt, a positive number.
•
It is an average: the secant through the two readings has slope Δ[A]/Δt, though the curve is steepest at the start.
Step-by-step
WE KNOW
[A]₁ = 1.2 mol/L, [A]₂ = 0.75 mol/L, t₁ = 30 s, t₂ = 120 s
FIND
time between (Δt), change in concentration (Δ[A]), average rate (r)
Step 1 · Find time between (Δt)
Use Δt = t₂ − t₁
The time between the two readings.
Δt = 120 − 30
Δt = 90 s
Step 2 · Find change in concentration (Δ[A])
Use Δ[A] = [A]₂ − [A]₁
Last minus first: negative, since the reactant is used up.
Δ[A] = 0.75 − 1.2
Δ[A] = −0.45 mol/L
Step 3 · Find average rate (r)
Use r = −Δ[A]/Δt
The drop in concentration per second: the minus sign makes the rate positive.
r
=
−(−0.45)/90
r
=
0.005
mol/(L·s)
Check
Put the numbers back in. Both sides should match.
90 = 120 − 30 ✓
−0.45 = 0.75 − 1.2 ✓
0.005 = −(−0.45)/90 ✓
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