ACTIVATION ENERGY Energy Reactants Products Reaction Progress

Energy EA Reactants DH Products Reaction Progress

Energy Reactants DH Products Reaction Progress

Energy Reactants DH Products Reaction Progress

Energy EA Reactants DH Products Reaction Progress

What happens here? Energy EA Reactants DH Products Reaction Progress

Effect of catalyst Without catalyst EA DH Reactants Products Energy EA Reactants DH Products Reaction Progress

Effect of catalyst With catalyst DH Reactants Products Energy With catalyst Reactants DH Products Reaction Progress

Effect of catalyst EA With catalyst DH Reactants Products Energy EA With catalyst Reactants DH Products Reaction Progress

Effect of catalyst Without catalyst EA DH Reactants Products Energy EA Reactants DH Products Reaction Progress

Effect of catalyst EA With catalyst DH Reactants Products Energy EA With catalyst Reactants DH Products Reaction Progress

Effect of catalyst Without catalyst EA EA With catalyst DH Reactants Energy EA EA With catalyst Reactants DH Products Reaction Progress

k = kNaOH = kHCl = kNaCl = kH2O ACTIVATION ENERGY Arrhenius equation The algebraic equation that relates – rA to the species concentrations is called the kinetic expression or as you know rate law. 1NaOH + 1HCl 1NaCl + 1H2O k = kNaOH = kHCl = kNaCl = kH2O It was the great Swedish chemist Arrhenius who first suggested that the temperature dependence of the specific reaction rate, k, could be correlated by an equation of the type: Where: A = pre-exponential factor, Ea = activation energy, J/mol or cal/mol, R = gas constant = 8.3 14 J/mol.K = 1.987 cal/mol.K, T = absolute temperature (K).

ACTIVATION ENERGY y = mx +b ln(x) = 2.303log(x) ln k = − Ea R 1 T Arrhenius equation After taking the natural logarithm of Equation: ln(x) = 2.303log(x) ln k = − Ea R 1 T + ln A y = mx +b

Arrhenius Equation y = m x + b ln k = − Ea R 1 T + ln A Fig 14.17 Graphical determination of activation energy Plot of ln k vs 1/T Slope = −Ea/R

The Arrehenius equation can be used to relate rate constants k1 and k2 at temperatures T1 and T2. ln k1 = − Ea R 1 T1 + ln A combine to give: ln k2 = − Ea R 1 T2 + ln A

1NaOH + 1HCl 1NaCl + 1H2O R = 1.9858775×10−3 kcal/mol.K T (ºC) k 20 0.021 30 0.052 40 0.171 50 0.306 60 0.319 T (K) 1/T ln k 293 0.003413 -3.86323 303 0.0033 -2.95651 313 0.003195 -1.76609 323 0.003096 -1.18417 333 0.003003 -1.14256 R = 1.9858775×10−3 kcal/mol.K

Example: The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K?

Solution We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k.

A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17: slope = − Ea/R = − 1.9 x 104

We use the value for the molar gas constant R in units of J/mol-K We use the value for the molar gas constant R in units of J/mol-K. We thus obtain

(b) To determine the rate constant, k1, at T1 = 430 (b) To determine the rate constant, k1, at T1 = 430.0 K, we can use Equation 14.21 with Ea = 160 kJ/mol, and one of the rate constants and temperatures from the given data, such as: k2 = 2.52 × 10–5 s–1 and T2 = 462.9 K: Thus, Note that the units of k1 are the same as those of k2.