Ch 13 4: Collision Theory; Activation Energy Temperature Dependence of Rate Constants • Collision Theory o Extension of kinetic molecular theory
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Ch. 13.4: Collision Theory; Activation Energy & Temperature Dependence of Rate Constants • Collision Theory o Extension of kinetic molecular theory § Reaction Rate ∝ number of collisions/sec § Reaction Rate increases w/increasing (absolute) temperature § Reaction Rate increases w/use of a catalyst § Reaction Rate increases w/increasing surface area (smaller particle size) § Reaction Rate increases w/increasing reactivity (e.g. Li < Na < K) o Effective Collisions § Only a small fraction of collisions result in reaction • Need sufficient energy & correct orientation • Orientation Factor: molecules must be oriented properly to achieve transition state (right). o Transition State § Transition state (activated complex [A⋯B]‡) is species intermediate between reactants & product • Reactant bonds breaking as product bonds forming (SN2 reaction from Organic, at right) § Can't collect activated complex, but can see evidence spectroscopically (bonds weakening, strengthening) o Activation Energy, Ea: minimum kinetic energy to achieve activated: § Energy needed to begin breaking reactant bonds and form activated complex (AB‡) • Ea for reverse reaction (endothermic): Ea,f + ∆H § Increasing temperature increases reaction rate because there are more molecules with enough kinetic energy § Decreasing Ea would also increase reaction rate • Factors in reaction rate o Concentration Effects: below left § Increasing concentration increases collisions § Collision Rate ∝ [A][B] • Doubling [A] or [B] doubles # of collisions, doubles effective collisions (reaction 1st order in A and in B) • Doubling [A] & [B] quadruples # of collisions & effective collisions • If reaction requires 2 A molecules to collide, effective rate depends on [A][A] = [A]2 or 2nd order in A o Doubling [A] quadruples rate Kinetics Kinetics ΔH
Collision Theory; Activation Energy & Temperature Dependence of Rate Constants p.2 • Temperature Effects (right) o Increasing temperature increases reaction rate • More collisions with enough kinetic energy, so more collisions are effective • Secondary effect: more collisions since particles moving faster o Catalysts speed up reaction • Decreasing Ea so more molecules have sufficient E • Mechanisms next lesson • Surface Area o Increasing surface area increases rate § Reactants more accessible § Decrease particle size (below right): • The Arrhenius Equation o Describes the effect of orientation, temperature, and activation energy on the on the reaction rate constant:
k = Ae -E a /RT, where A is the frequency factor (# collisions w/correct orientation, assumed independent of T), Ea is in J/mol, T in kelvins, and R=8.31 J/mol⋅K § As Ea increases, exponential term gets smaller, so k decreases and reaction rate decreases o Taking ln of both side of the Arrhenius Equation: , a straight-line plot of lnk vs. 1/T, with slope = -Ea/R and y-intercept = lnA § Determine the activation energy for the isomerization of methylisonitrile to acetonitrile, CH3NC → CH3CN from the following determinations of k at various temperatures (fill in last 3 columns): • Plot of lnk vs. 1/T (in K-1) gives slope = -1.91×104 K:
E a = -slope × R = -(-1.91×10 4 K)8.314 J
mol⋅K 1 kJ1000 J
= 159 kJ/mol § Use an alternate form comparing k at two temperatures:Slope=
Δy Δx lnk 2 -lnk 1 T 2 -1 -T 1 -1 -E a R , so lnk 2 -lnk 1 = ln k 2 k 1 -E a R 1 T 2 1 T 1 • For the same reaction above, what is the value of k when the temperature is 430.0 K? ln k 2 k 1 -E a R 1 T 2 1 T 1 : ln k 22.52×10
-5 s -1 -159 kJ/mol8.314 J/mol⋅K
1000 J
1 kJ 1430.0 K
1462.9 K
ln k 22.52×10
-5 s -1 =-3.16, so k 2 =(2.52×10 -5 s -1 )e -3.16 =1.07×10 -6 s -1Temperature, °C k, s-1 189.7 2.52×10-5 198.9 5.25×10-5 230.3 6.30×10-4 251.2 3.16×10-3
lnk = -E a R 1 T + lnACollision Theory; Activation Energy & Temperature Dependence of Rate Constants p.3 § This could also be used to determine Ea from two values of k at two temperatures: • A first-order reaction has rate constants k1 = 4.6×10-2 s-1 and k2 = 8.1×10-2 s-1 at 0°C and 20°C, respectively. What is the value of the activation energy?
ln k 2 k 1 -E a R 1 T 2 1 T 1 ;ln 8.1 4.6 -E a8.31 J/mol⋅K
1 293 K1 273 K
E a =8.31J/mol 0.566