Enter at least two sets of temperature and rate constant data.
The calculator converts the values to (1/T) × 10³
and ln k, plots the Arrhenius graph and determines
the activation energy from its gradient.
Set
Temperature
Rate constant, k
(1/T) × 10³
ln k
The normal graph focuses on the experimental data. Select the option
above to extend the best-fit line to 1/T = 0 and
show the graphical meaning of ln A.
Gradient
—
Activation energy, Eₐ
—
Y-intercept = ln A
—
Pre-exponential factor, A
—
R² value
—
Best-fit equation
—
Reading the Arrhenius graph
ln k = −(Eₐ/R)(1/T) + ln A
Because the x-axis uses (1/T) × 10³:
gradient → activation energy, Eₐ
Eₐ = −gradient × R in kJ mol⁻¹
y-intercept → ln A
A = eintercept
The pre-exponential factor A has the same units as
the rate constant k. These units depend upon the
rate equation.
Explore Temperature and Activation Energy
Explore how changing the temperature or the
activation energy affects the rate constant,
k. The starting conditions are defined as having
a relative rate constant of 1.00.
Reference conditions
At these reference conditions, k₂/k₁ = 1.00.
Raise or lower the temperature
K
Activation energy is held constant while the temperature changes.
kJ mol⁻¹
Temperature is held constant while the activation energy changes.
Relative rate constant
k₂ / k₁
1.00
The rate constant is unchanged.
The vertical axis is logarithmic. Each major step represents a
tenfold change in the rate constant.
Why does increasing temperature increase k?
Increasing the temperature makes the exponential term in the
Arrhenius equation less negative. The value of k therefore increases.
Notice that the relationship is not linear.
ln(k₂/k₁) = −(Eₐ/R)(1/T₂ − 1/T₁)
A catalyst increases the rate constant by providing an alternative
reaction pathway with a lower activation energy. It does not increase
the temperature of the reaction.