Calculate concentration, time, and rate constants for different reaction orders
Calculate concentration, time, or rate constant for zero, first, and second order reactions.
Zero Order:
Concentration decreases linearly with time. Common in surface-catalyzed reactions and enzyme kinetics when substrate is saturated. Half-life decreases as reaction proceeds.
First Order:
Concentration decreases exponentially with time. Common in radioactive decay and many decomposition reactions. Half-life is constant (independent of concentration).
Second Order:
1/[A] increases linearly with time. Common in gas-phase reactions and dimerizations. Half-life increases as reaction proceeds (takes longer when concentration is lower).
Graphical Analysis:
Integrated rate laws are mathematical equations that relate the concentration of a reactant to time. Unlike differential rate laws (which express rate as a function of concentration), integrated rate laws allow us to predict concentrations at any point in time.
The form of the integrated rate law depends on the order of the reaction with respect to the reactant. The three most common cases are:
Key Concept:
Integrated rate laws are derived by integrating the differential rate law. They provide a direct relationship between concentration and time, making them extremely useful for experimental kinetics and predicting reaction progress.
Differential Rate Law:
Rate = k
Integrated Rate Law:
[A] = [A]₀ - kt
(Linear decrease in concentration with time)
Half-Life:
t½ = [A]₀ / (2k)
(Decreases as reaction proceeds - depends on initial concentration)
Graphical Analysis:
Plot [A] vs t → straight line with slope = -k
Examples:
Differential Rate Law:
Rate = k[A]
Integrated Rate Law:
ln[A] = ln[A]₀ - kt
or [A] = [A]₀e^(-kt)
(Exponential decrease in concentration with time)
Half-Life:
t½ = ln(2) / k = 0.693 / k
(Constant - independent of concentration!)
Graphical Analysis:
Plot ln[A] vs t → straight line with slope = -k
Examples:
Differential Rate Law:
Rate = k[A]²
Integrated Rate Law:
1/[A] = 1/[A]₀ + kt
(Reciprocal of concentration increases linearly with time)
Half-Life:
t½ = 1 / (k[A]₀)
(Increases as reaction proceeds - depends on initial concentration)
Graphical Analysis:
Plot 1/[A] vs t → straight line with slope = k
Examples:
| Property | Zero Order | First Order | Second Order |
|---|---|---|---|
| Rate Law | Rate = k | Rate = k[A] | Rate = k[A]² |
| Integrated Form | [A] = [A]₀ - kt | ln[A] = ln[A]₀ - kt | 1/[A] = 1/[A]₀ + kt |
| Half-Life | [A]₀/(2k) | 0.693/k | 1/(k[A]₀) |
| Linear Plot | [A] vs t | ln[A] vs t | 1/[A] vs t |
| Slope of Plot | -k | -k | +k |
| Units of k | M·s⁻¹ | s⁻¹ | M⁻¹·s⁻¹ |
| t½ Dependence | Decreases | Constant | Increases |
The decomposition of N₂O₅ is first order with k = 5.7 × 10⁻⁴ s⁻¹. If the initial concentration is 0.100 M, what will be the concentration after 2000 seconds?
Step 1: Write the integrated rate law for first order
[A] = [A]₀ e^(-kt)
Step 2: Substitute values
[A] = (0.100 M) × e^(-(5.7×10⁻⁴ s⁻¹)(2000 s))
Step 3: Calculate the exponent
-kt = -(5.7×10⁻⁴)(2000) = -1.14
Step 4: Calculate e^(-1.14)
e^(-1.14) = 0.3198
Step 5: Calculate final concentration
[A] = (0.100 M)(0.3198) = 0.0320 M
Answer:
After 2000 seconds, the concentration will be 0.0320 M, which is about 32% of the original concentration.
Note: The half-life for this reaction is t½ = 0.693/k = 0.693/(5.7×10⁻⁴) = 1216 s. Since 2000 s is slightly more than one half-life, having about 32% remaining (slightly less than 50%) makes sense.
Most drug metabolism follows first-order kinetics. Integrated rate laws help determine dosing schedules, predict drug concentrations in blood over time, and calculate how long it takes for a drug to be eliminated from the body.
Radioactive decay is first-order. Carbon-14 dating uses the integrated rate law to determine the age of archaeological artifacts. By measuring remaining C-14 and knowing its half-life (5,730 years), we can calculate the age of organic materials.
Degradation of pollutants often follows first-order kinetics. Integrated rate laws predict how long it takes for pesticide residues, pharmaceutical contaminants, or oil spills to degrade to safe levels in soil and water.
Food spoilage reactions (vitamin degradation, flavor loss, color changes) often follow first-order kinetics. Integrated rate laws help determine shelf life, optimal storage conditions, and expiration dates for food products.
Industrial reactor design requires integrated rate laws to determine reactor size, residence time, and optimal operating conditions. Engineers use these equations to scale up processes from laboratory to production scale.
Ozone depletion, greenhouse gas reactions, and smog formation all involve reactions with specific rate laws. Integrated rate laws help model atmospheric changes and predict future concentrations of important atmospheric species.
The integrated rate law for first order uses natural logarithm (ln), not log base 10.
Correct: ln[A] = ln[A]₀ - kt
After calculating ln[A], you must use e^x to get back to [A].
Correct: If ln[A] = -1.5, then [A] = e^(-1.5) = 0.223, not -1.5
For zero order: [A] = [A]₀ minus kt (not plus). For second order: 1/[A] = 1/[A]₀ plus kt.
Remember: Concentration always decreases, but the mathematical form varies by order.
If k is in s⁻¹, time must be in seconds. If k is in min⁻¹, time must be in minutes.
Correct approach: Convert all units to match before calculating.
[A]₀ is always the initial concentration (at t=0). [A] is the concentration at time t.
Remember: The subscript 0 means "initial" or "at time zero".
Using a first-order equation for a second-order reaction (or vice versa) will give completely wrong results.
Correct approach: Always verify the reaction order before choosing the equation.
[A] = [A]₀ - kt
t½ = [A]₀/(2k)
Plot [A] vs t → line
Slope = -k
Units: M·s⁻¹
[A] = [A]₀e^(-kt)
t½ = 0.693/k
Plot ln[A] vs t → line
Slope = -k
Units: s⁻¹
1/[A] = 1/[A]₀ + kt
t½ = 1/(k[A]₀)
Plot 1/[A] vs t → line
Slope = k
Units: M⁻¹·s⁻¹