Calculate rate constants and understand temperature effects on reaction rates
Arrhenius Equation: k = A·e^(-Ea/RT)
Relates rate constant to temperature and activation energy
Higher temperature or lower activation energy → faster reaction
kJ/mol
Same units as k (use scientific notation, e.g., 1e13)
K (Kelvin)
k = A·e^(-Ea/RT)
ln(k) = ln(A) - Ea/(RT)
ln(k₂/k₁) = -(Ea/R)·(1/T₂ - 1/T₁)
The Arrhenius equation describes how the rate constant (k) of a chemical reaction depends on temperature and activation energy. It was proposed by Swedish chemist Svante Arrhenius in 1889.
k = A · e^(-Ea/RT)
The proportionality constant in the rate law that relates reaction rate to reactant concentrations.
• Larger k → faster reaction
• Increases exponentially with temperature
• Units: s⁻¹ (first order), M⁻¹s⁻¹ (second order), etc.
The minimum energy required for reactants to form products. Energy barrier to overcome.
• Higher Ea → slower reaction (harder to activate)
• Lower Ea → faster reaction (easier to activate)
• Catalysts lower Ea without being consumed
• Typical range: 50-250 kJ/mol for most reactions
Also called the frequency factor. Related to collision frequency and orientation.
• Represents frequency of collisions with proper orientation
• Generally between 10⁸ and 10¹⁴ for most reactions
• Weakly temperature-dependent (often treated as constant)
• Higher A → more frequent successful collisions
Absolute temperature in Kelvin. Must be positive and affects k exponentially.
• Must use Kelvin: K = °C + 273.15
• Higher T → more molecules with E ≥ Ea
• Rule of thumb: k doubles every 10°C increase
• Exponential relationship with k
k = A·e^(-Ea/RT)
Most common form. Directly calculates k from Ea, A, and T.
ln(k) = ln(A) - Ea/(RT)
Linear form: y = mx + b
Plot ln(k) vs 1/T gives straight line with slope = -Ea/R and intercept = ln(A)
ln(k₂/k₁) = -(Ea/R)·(1/T₂ - 1/T₁)
Useful when you know k at two different temperatures. Eliminates need to know A.
Problem: A reaction has Ea = 100 kJ/mol and A = 1.0 × 10¹³ s⁻¹. What is k at 500 K?
k = A·e^(-Ea/RT)
k = (1.0×10¹³)·e^(-100,000/(8.314×500))
k = (1.0×10¹³)·e^(-24.06)
k = (1.0×10¹³)·(3.42×10⁻¹¹)
k ≈ 342 s⁻¹
Problem: k = 0.001 s⁻¹ at 298 K. If Ea = 75 kJ/mol, what is k at 323 K?
ln(k₂/k₁) = -(Ea/R)·(1/T₂ - 1/T₁)
ln(k₂/0.001) = -(75,000/8.314)·(1/323 - 1/298)
ln(k₂/0.001) = -9022·(-0.000260)
ln(k₂/0.001) = 2.346
k₂/0.001 = 10.44
k₂ ≈ 0.0104 s⁻¹ (about 10× faster!)
An Arrhenius plot is a graph of ln(k) vs 1/T. The logarithmic form of the Arrhenius equation produces a straight line.
ln(k) = -Ea/R · (1/T) + ln(A)
y = m·x + b
Always use Kelvin for temperature in the Arrhenius equation. Using Celsius or Fahrenheit will give incorrect results. Convert: K = °C + 273.15
The Arrhenius equation works best over moderate temperature ranges. At extreme temperatures, A and Ea may change, and the equation becomes less accurate. Physical state changes (melting, boiling) also affect applicability.
To experimentally determine Ea and A, measure k at several different temperatures, plot ln(k) vs 1/T, and calculate slope and intercept. Need at least 3-4 data points for reliable results.
Catalysts increase reaction rate by lowering Ea (not by changing A or T). This is why enzymes and industrial catalysts are so important - they make reactions feasible at lower temperatures.