Chemical Kinetics Temperature Dependence of Reaction Rates

This is the behaviour of k with temperature for many reactions. Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. “k versus T” This is the behaviour of k with temperature for many reactions. CHEM 3310

The curve can be fitted to a Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. The curve can be fitted to a y=aex relationship. CHEM 3310

“k versus Temperature” In 1887, Svante Arrhenius suggested that rate constants vary exponentially with the reciprocal of the absolute temperature. “k versus Temperature” “ln k versus 1/T” k = b eaT It turns out all reactions have rate constants with this equation, but with different constants a and b for each reaction. CHEM 3310

In 1887, Svante Arrhenius suggested that rate constants vary exponentially with the reciprocal of the absolute temperature. Arrhenius’ Equation (The above equation is purely empirical.) Note: R is the gas constant (8.314 J mole-1 K-1, 1.98 cal mole-1 K-1) T is the temperature in Kelvin Ea is the activation energy 2. RT is in units of energy per mole, thus, Ea is also in units of energy per mole. A has the same units as the rate constant, k. A is a number that represents the likelihood that collisions would occur in a reaction. A is the frequency factor CHEM 3310

Other forms of Arrhenius’ Equation 1. Take the natural logarithm of both sides of the equation Linear form y = m x + b When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of “ln k vs. 1/T” Slope = -Ea / R CHEM 3310

Other forms of Arrhenius’ Equation Example Determine the activation energy for the decomposition of N2O5 from the following temperature dependence rate constant data. k (s1) Temperature (°C) 4.8x104 45.0 8.8x104 50.0 1.6x103 55.0 2.8x103 60.0 N2O5 (g)  2 NO2 (g) + ½ O2 (g) Recall: Rate = k [N2O5] CHEM 3310

Other forms of Arrhenius’ Equation 2. For two temperatures, Subtract the two equations Two k and temperatures form Note: A drops out of the equation when it is in this form. CHEM 3310

Other forms of Arrhenius’ Equation Example Determine the rate constant at 35°C for the hydrolysis of sucrose, given that at 37°C it is 0.91 mL mole-1 sec-1. The activation energy of this reaction is 108 kJ/mol. Calculation shows that: k2/k1 = 1.31 An increase of 2oC increases the reaction rate by 31%. k1, the rate constant at 35oC is 0.00069 L mole-1 sec-1. Rate constant increases when T2>T1 k2 = 0.91 mL mole-1 sec-1 = 9.1 x 10-4 L mole-1 sec-1 T2 = 37oC = 310.1 K T1 = 35oC = 308.1 K Ea = 108 kJ mole-1 Calculate k1 CHEM 3310

Arrhenius’ Equation What else can we get from the slope of Arrhenius’ plot? High activation energy corresponds to a reaction rate that is very sensitive to temperature (slope of Arrhenius graph is steep) Slope = -Ea / R Small activation energy indicates a rate that varies only slightly with temperature. CHEM 3310

“k versus Temperature” Why does the temperature dependence obey the relationship shown below? “k versus Temperature” “ln k versus 1/T” k = b eaT Is there any physical insights into the order of the reaction or the meaning of the constants (Ea/R) and A in this dependency? CHEM 3310

Second Order Reactions Magnitudes of A and Ea First Order Reactions A (s-1) Ea kJ/mole) cyclopropene  propane 1.58 x 1015 272 2 N2O5  4 NO2 + O2 4.94 x 1013 103 N2O  N2 + O 7.94 x 1011 250 Second Order Reactions A (L mole-1s-1) Ea kJ/mole) Sucrose in acidic water 1.50 x 1015 107.9 CHEM 3310

As Ea increases, k decreases. Summary Arrhenius Equation In natural logarithmic form, y = m x + b Determine Ea from the slope. Determine A from the intercept. As Ea increases, k decreases. CHEM 3310