Temperature Dependence & the Arrhenius Equation

Why Temperature Speeds Up Reactions

Almost every reaction goes faster when heated. A useful rule of thumb is: the rate roughly doubles for every 10 K rise in temperature. Why such a large effect from a modest temperature change?

It is not mainly that molecules move a little faster. The real reason is that raising the temperature sharply increases the fraction of molecules that have enough energy to react (the activation energy). This is captured quantitatively by the Arrhenius equation.

$$\boxed{k = A\,e^{-E_a/RT}}$$

• $k$ = rate constant,
• $A$ = the Arrhenius factor or frequency factor (related to collision frequency and orientation),
• $E_a$ = activation energy (J mol$^{-1}$),
• $R$ = gas constant (8.314 J K$^{-1}$ mol$^{-1}$),
• $T$ = absolute temperature (K).

The Logarithmic Forms

Taking natural logs of the Arrhenius equation:

$$\ln k = \ln A - \frac{E_a}{RT}$$

So a plot of $\ln k$ against $1/T$ is a straight line with slope $-\dfrac{E_a}{R}$ and intercept $\ln A$. In base-10:

$$\log k = \log A - \frac{E_a}{2.303\,RT}$$

Two-temperature form (for rate constants $k_1$ at $T_1$ and $k_2$ at $T_2$):

$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

This is the workhorse formula for finding $E_a$ from two rate constants, or one $k$ given the other.

Activation Energy and the Energy Barrier

For a reaction to occur, colliding molecules must first form an unstable, high-energy activated complex (transition state) at the top of an energy barrier. The minimum extra energy reactants need to reach that barrier is the activation energy $E_a$.

• A high $E_a$ means few molecules have enough energy → slow reaction.
• A low $E_a$ means many molecules can react → fast reaction.

The Arrhenius factor $e^{-E_a/RT}$ is the fraction of molecules that have energy at least $E_a$ at temperature $T$ in the simplified Arrhenius picture. Raising $T$ or lowering $E_a$ increases this fraction and hence the rate.

Key Point: From $k = A e^{-E_a/RT}$, both increasing temperature and decreasing activation energy (e.g. by a catalyst) increase the rate constant. The exponential dependence is why a 10 K rise can nearly double the rate.

[JEE Tip] When using the two-temperature formula, keep $E_a$ in J mol$^{-1}$ with $R = 8.314$, or in kJ mol$^{-1}$ with $R = 8.314\times10^{-3}$. Temperatures must be in kelvin. Mismatched units are a common source of error.

Solved Examples

Example 1: Rate doubling rule

The rate of a reaction doubles when temperature rises from 298 K to 308 K. Estimate $E_a$. ($R = 8.314$ J K$^{-1}$ mol$^{-1}$)

Solution: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\dfrac{T_2 - T_1}{T_1 T_2}$. $\log 2 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{10}{298\times308}$. $0.301 = \dfrac{E_a}{19.147}\times\dfrac{10}{91784}$. $E_a = \dfrac{0.301\times19.147\times91784}{10} = 52897$ J $\approx 52.9$ kJ mol$^{-1}$.

Example 2: Slope of an Arrhenius plot

An Arrhenius plot of $\ln k$ vs $1/T$ has slope $-6000$ K. Find $E_a$. ($R = 8.314$)

Solution: slope $= -E_a/R$, so $E_a = -\text{slope}\times R = 6000\times8.314 = 49884$ J $\approx 49.9$ kJ mol$^{-1}$.

Example 3: Finding k at another temperature

A reaction has $k_1 = 2.0\times10^{-3}$ s$^{-1}$ at 300 K and $E_a = 50$ kJ mol$^{-1}$. Find $k_2$ at 310 K. ($R = 8.314$)

Solution: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\dfrac{T_2-T_1}{T_1T_2} = \dfrac{50000}{2.303\times8.314}\times\dfrac{10}{300\times310}$. $= 2613.3\times\dfrac{10}{93000} = 2613.3\times1.075\times10^{-4} = 0.281$. So $\dfrac{k_2}{k_1} = 10^{0.281} = 1.91$, giving $k_2 = 1.91\times2.0\times10^{-3} = 3.82\times10^{-3}$ s$^{-1}$.

Example 4: Activation energy from two rate constants

For a reaction, $k$ increases from $1.0\times10^{-3}$ to $4.0\times10^{-3}$ s$^{-1}$ when $T$ rises from 300 K to 320 K. Find $E_a$. ($R = 8.314$)

Solution: $\log\dfrac{4.0\times10^{-3}}{1.0\times10^{-3}} = \log4 = 0.602$. $0.602 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{20}{300\times320}$. $0.602 = \dfrac{E_a}{19.147}\times\dfrac{20}{96000}$. $E_a = \dfrac{0.602\times19.147\times96000}{20} = 55315$ J $\approx 55.3$ kJ mol$^{-1}$.

Example 5: Meaning of the frequency factor

In $k = A e^{-E_a/RT}$, what does $A$ represent?

Solution: $A$ is the frequency (or Arrhenius) factor — it reflects the frequency of collisions and the fraction with proper orientation. It is approximately the rate constant the reaction would have if the activation barrier were negligible and collisions were effectively successful.

Example 6: Effect of catalyst on Arrhenius terms

How does a catalyst change $k$ in the Arrhenius equation?

Solution: A catalyst lowers the activation energy $E_a$. Since $k = A e^{-E_a/RT}$, a smaller $E_a$ makes the exponential term larger, so $k$ (and the rate) increases. For a given temperature, $A$ is usually taken as unchanged in this simplified treatment.

Example 7: Fraction of molecules above $E_a$

The term $e^{-E_a/RT}$ represents what physical quantity?

Solution: It is the fraction of molecules whose kinetic energy is equal to or greater than the activation energy $E_a$ at temperature $T$ in the Arrhenius model. Raising $T$ increases this fraction sharply.

Example 8: Temperature for a target rate increase

If $E_a = 0$, what does the Arrhenius equation predict about temperature dependence?

Solution: With $E_a = 0$, $k = A e^{0} = A$, a constant. The rate constant would be independent of temperature in this idealized case.

Example 9: Two-temperature numerical

A first-order reaction has $k = 1.0\times10^{-2}$ s$^{-1}$ at 400 K. If $E_a = 80$ kJ mol$^{-1}$, find $k$ at 420 K. ($R = 8.314$)

Solution: $\log\dfrac{k_2}{k_1} = \dfrac{80000}{2.303\times8.314}\times\dfrac{20}{400\times420} = 4181.3\times\dfrac{20}{168000} = 4181.3\times1.190\times10^{-4} = 0.498$. $\dfrac{k_2}{k_1} = 10^{0.498} = 3.15$, so $k_2 = 3.15\times10^{-2}$ s$^{-1}$.

Example 10: Why rate rises faster than molecular speed

A 10 K rise increases average molecular speed by only ~2%, yet the rate can double. Explain.

Solution: Rate depends not on average speed but on the fraction of molecules exceeding $E_a$, given by $e^{-E_a/RT}$. This fraction is very sensitive to temperature, so a small $T$ rise produces a large rate increase.