Chpt 12 - Chemical Kinetics. Reaction Rates Rate Laws Reaction Mechanisms Collision Theory Catalysis HW set1: Chpt 12 - pg. 580-592, # 22, 23, 28 Due Dec. 13 HW set2: Chpt 12 # 39, 48, 53, 56, 58, 60 due Dec. 17 HW set3: Chpt 12 #62, 69, 72 due Dec. 19. Reaction Rate ? .
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[A] means concentration of A in mol/L; A is the reactant or product being considered.
2NO2 --> 2NO + O2
2NO2 --> 2NO + O2
The rate of comsumption (disappearance) of NO2 is same as rate of production of NO and twice the rate of production of O2.
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n:
Rate = k[NO2]n
Rate = k[A]n[B]m
Overall reaction order = n + m
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
Take actual data for each reactant and plot it… k is slope of line
[A] vs t or ln[A] vs t or 1/[A] vs t straight line says Oo, 1o, 2o
for calculations at any time t
t = time, [A]o = initial conc, [A] = conc at t, k = rate constant
A plot of [A] vs time
Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is:
A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
Elementary Steps (Molecularity)
2N2O5(g) 4NO2(g) + O2(g)
Step 1: 2( N2O5 NO2 + NO3 ) (fast)
Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)
Step 3: NO3 + NO → 2NO2 (fast)
Does this satisfy overall balanced eqn?
The reaction A + 2B C has the following proposed mechanism:
A + B D (fast equilibrium)
D + B C (slow)
Write the rate law for this mechanism.
rate = k[A][B]2
What if the fast and slow were reversed?
How fast a reaction happens is related by energy
Number of collisions with Ea = (total collisions) x e-Ea/RT
where R is gas constant and T is temp (K)
But this value was still too high for observed rates so total collisions must not all react (orientation effect)
so k = zpe-Ea/RT with z = collision frequency and p = steric factor (orientation)
A = frequency factor (zp)
Ea = activation energy
R = gas constant (8.3145 J/K·mol)
T = temperature (in K)
Slope = ?
Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C?
Ea = 53 kJ
The catalyst provides a new pathway which has a lower Ea
It does not lower the Ea for the original rxn
Catalyzed Rxn has a lower Ea, so many more collisions are effective at making the product(s)