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Chemical Kinetics - Activation energy (HL)

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Activation energy (EaE_a) is the minimum energy that colliding particles must possess for a reaction to occur upon collision.

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The temperature dependence of the rate constant is expressed by the Arrhenius equation: k=Ae−EaRTk = A e^{-\frac{E_a}{RT}}.

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The pre-exponential factor (AA), also known as the frequency factor, accounts for the frequency of collisions and the probability that they occur with the correct orientation (steric factor).

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A small increase in temperature (TT) leads to a significant increase in the rate constant (kk) because the fraction of molecules with E≥EaE \ge E_a increases exponentially.

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The Arrhenius equation can be expressed in logarithmic form: ln⁡k=−EaR(1T)+ln⁡A\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A. This is a linear equation of the form y=mx+cy = mx + c.

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On an Arrhenius plot of ln⁡k\ln k against 1T\frac{1}{T}, the gradient (mm) of the line is equal to −EaR-\frac{E_a}{R}.

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Catalysts provide an alternative reaction pathway with a lower EaE_a. This increases the value of kk without changing the temperature.

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The gas constant RR is 8.31 J K−1 mol−18.31 \text{ J K}^{-1} \text{ mol}^{-1}. Ensure units for EaE_a are converted from kJ mol−1\text{kJ mol}^{-1} to J mol−1\text{J mol}^{-1} when using this value.

📐Formulae

k=Ae−EaRTk = A e^{-\frac{E_a}{RT}}

ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}

ln⁡(k1k2)=EaR(1T2−1T1)\ln \left(\frac{k_1}{k_2}\right) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)

gradient=−EaR\text{gradient} = -\frac{E_a}{R}

💡Examples

Problem 1:

A reaction has a rate constant k1=4.5×10−3 s−1k_1 = 4.5 \times 10^{-3} \text{ s}^{-1} at 298 K298 \text{ K}. When the temperature is increased to 313 K313 \text{ K}, the rate constant becomes k2=1.8×10−2 s−1k_2 = 1.8 \times 10^{-2} \text{ s}^{-1}. Calculate the activation energy (EaE_a) for this reaction in kJ mol−1\text{kJ mol}^{-1}.

Solution:

ln⁡(4.5×10−31.8×10−2)=Ea8.31(1313−1298)\ln \left(\frac{4.5 \times 10^{-3}}{1.8 \times 10^{-2}}\right) = \frac{E_a}{8.31} \left(\frac{1}{313} - \frac{1}{298}\right) ln⁡(0.25)=Ea8.31(0.003195−0.003356)\ln(0.25) = \frac{E_a}{8.31} \left(0.003195 - 0.003356\right) −1.386=Ea8.31(−1.61×10−4)-1.386 = \frac{E_a}{8.31} \left(-1.61 \times 10^{-4}\right) Ea=−1.386×8.31−1.61×10−4=71538 J mol−1E_a = \frac{-1.386 \times 8.31}{-1.61 \times 10^{-4}} = 71538 \text{ J mol}^{-1} Ea≈71.5 kJ mol−1E_a \approx 71.5 \text{ kJ mol}^{-1}

Explanation:

The two-point form of the Arrhenius equation is used to solve for EaE_a. Note that temperature must be in Kelvin and the final answer is converted to kJ mol−1\text{kJ mol}^{-1} by dividing by 10001000.

Problem 2:

The slope of a graph of ln⁡k\ln k vs 1T\frac{1}{T} for a specific reaction is −6.5×103 K-6.5 \times 10^3 \text{ K}. Determine the activation energy of the reaction.

Solution:

gradient=−EaR\text{gradient} = -\frac{E_a}{R} −6.5×103=−Ea8.31-6.5 \times 10^3 = -\frac{E_a}{8.31} Ea=6.5×103×8.31=54015 J mol−1E_a = 6.5 \times 10^3 \times 8.31 = 54015 \text{ J mol}^{-1} Ea=54.0 kJ mol−1E_a = 54.0 \text{ kJ mol}^{-1}

Explanation:

In an Arrhenius plot, the gradient is defined as −EaR-\frac{E_a}{R}. By multiplying the absolute value of the slope by the gas constant RR, we find the activation energy in J mol−1\text{J mol}^{-1}.