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Arrhenius Equation & Activation Energy cheat sheet - grade 11-12

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Chemistry Grade 11-12

Arrhenius Equation & Activation Energy Cheat Sheet

A printable reference covering the Arrhenius equation, activation energy, rate constants, temperature effects, and Arrhenius plots for grades 11-12.

TopicsChemical ReactionsEnergyThermodynamicsGraphsChemistryStudy Guide
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The Arrhenius equation explains how temperature affects the rate of a chemical reaction. This cheat sheet helps students connect particle collisions, activation energy, and rate constants in one clear reference. It is especially useful for interpreting lab data, comparing reactions, and solving chemistry problems involving temperature changes.

Students need it because small temperature changes can cause large changes in reaction rate.

The core equation is k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}, where kk is the rate constant, AA is the frequency factor, EaE_a is activation energy, RR is the gas constant, and TT is temperature in kelvins. The linear form, ln⁡k=−EaR(1T)+ln⁡A\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A, allows students to find activation energy from a graph. A two-temperature form, ln⁡(k2k1)=−EaR(1T2−1T1)\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right), compares rate constants at two temperatures.

Higher activation energy usually means a reaction is more sensitive to temperature changes.

Key Facts

  • The Arrhenius equation is k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}, where kk is the rate constant and TT must be measured in kelvins.
  • Activation energy EaE_a is the minimum energy particles must have for a successful reaction to occur.
  • The gas constant is commonly used as R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} when EaE_a is measured in joules per mole.
  • The linear Arrhenius form is ln⁡k=−EaR(1T)+ln⁡A\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A.
  • On a graph of ln⁡k\ln k versus 1T\frac{1}{T}, the slope is m=−EaRm = -\frac{E_a}{R}, so Ea=−mRE_a = -mR.
  • The two-point Arrhenius equation is ln⁡(k2k1)=−EaR(1T2−1T1)\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).
  • Temperature must be converted using TK=T∘C+273.15T_{K} = T_{^{\circ}\text{C}} + 273.15 before using any Arrhenius equation.
  • A catalyst lowers EaE_a, which increases kk at the same temperature without changing the overall reaction energy difference.

Vocabulary

Arrhenius Equation
An equation, k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}, that relates a reaction rate constant to temperature and activation energy.
Activation Energy
The minimum energy, EaE_a, that reacting particles must have to form products successfully.
Rate Constant
The value kk that connects reactant concentration to reaction rate for a specific reaction at a specific temperature.
Frequency Factor
The value AA that represents how often particles collide with the proper orientation for reaction.
Arrhenius Plot
A graph of ln⁡k\ln k versus 1T\frac{1}{T} used to determine activation energy from the slope.
Catalyst
A substance that increases reaction rate by providing a lower-energy pathway and reducing EaE_a.

Common Mistakes to Avoid

  • Using Celsius instead of kelvins is wrong because the Arrhenius equation requires absolute temperature, so always convert with TK=T∘C+273.15T_{K} = T_{^{\circ}\text{C}} + 273.15.
  • Mixing joules and kilojoules is wrong because R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} requires EaE_a in J mol−1\text{J mol}^{-1}, not kJ mol−1\text{kJ mol}^{-1}.
  • Forgetting the negative slope is wrong because an Arrhenius plot has slope m=−EaRm = -\frac{E_a}{R}, so EaE_a must be calculated as Ea=−mRE_a = -mR.
  • Using log⁡\log instead of ln⁡\ln without conversion is wrong because the standard Arrhenius forms use natural logarithms, not base-10 logarithms.
  • Assuming a catalyst changes the products is wrong because a catalyst lowers EaE_a and speeds the reaction without changing the balanced equation or overall energy change.

Practice Questions

  1. 1 A reaction has Ea=75.0 kJ mol−1E_a = 75.0\ \text{kJ mol}^{-1} and A=2.5×1012 s−1A = 2.5 \times 10^{12}\ \text{s}^{-1}. Calculate kk at T=298 KT = 298\ \text{K} using k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}.
  2. 2 For an Arrhenius plot of ln⁡k\ln k versus 1T\frac{1}{T}, the slope is −9500 K-9500\ \text{K}. Calculate EaE_a in kJ mol−1\text{kJ mol}^{-1} using Ea=−mRE_a = -mR.
  3. 3 A reaction has k1=0.015 s−1k_1 = 0.015\ \text{s}^{-1} at T1=300 KT_1 = 300\ \text{K} and Ea=48.0 kJ mol−1E_a = 48.0\ \text{kJ mol}^{-1}. Use ln⁡(k2k1)=−EaR(1T2−1T1)\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) to find k2k_2 at T2=330 KT_2 = 330\ \text{K}.
  4. 4 Explain why a reaction with a larger EaE_a usually shows a greater increase in rate when temperature rises.

Understanding Arrhenius Equation & Activation Energy

Particles in a sample do not all move with the same energy. At any instant, some have little kinetic energy while a smaller group has much more. Heating shifts the whole energy distribution toward higher values.

The important change is not only that particles move faster. A much larger fraction can reach or pass the energy barrier for the reaction.

This explains why a modest rise in temperature can produce a noticeable rate change. The effect is strongest when the barrier is high, because the starting fraction of particles with enough energy is very small.

Activation energy represents the energy needed to reach a temporary, unstable arrangement of atoms called the transition state. Bonds may be stretching, breaking, or beginning to form in this arrangement. Reaching the transition state does not guarantee that every collision makes products.

Particles must approach with a suitable orientation as well as enough energy. The frequency factor accounts for features such as collision frequency and orientation.

Its value can change somewhat with temperature, but activation energy usually has the larger effect in school level calculations. Different reaction pathways can have different barriers, even when they begin with the same reactants and end with the same products.

A catalyst provides an alternative pathway with a lower barrier. It does not give particles extra energy and it is not used up overall. Instead, it may hold reactants in a useful position, weaken a bond, or form short lived intermediate substances.

Enzymes in living cells are catalysts of this kind. They allow reactions involved in digestion and respiration to happen quickly at body temperature.

Catalysts matter in car exhaust systems, fertilizer production, food processing, and industrial chemistry. They change how quickly equilibrium is reached, but they do not change the equilibrium position or the total energy change between reactants and products.

Arrhenius plots turn rate data into a straight line so that a hidden energy barrier can be estimated. Careful graph work matters. Use temperature in kelvins before finding reciprocal temperature.

Put the natural logarithm of the rate constant on the vertical axis and reciprocal temperature on the horizontal axis. The line slopes downward because reciprocal temperature decreases as temperature rises while the rate constant rises. Check units before multiplying the slope by the gas constant.

If the gas constant uses joules per mole per kelvin, the activation energy comes out in joules per mole. Experimental points may not form a perfect line because of measurement uncertainty, changing reaction conditions, or a reaction mechanism that changes across the temperature range. In labs, keep concentrations and catalysts constant while testing temperature, otherwise the measured change cannot be assigned to temperature alone.