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Lecture

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Department
Chemistry
Course Code
CHM135H1
Professor
Kris Quinlan

Page:
of 3
LECTURE 8: Kinetics and Equilibrium
Slide No. Notes
KINETICS
31 -lower to higher temperature, energy of collisions increases
-for a reaction to occur, there must be a collision and the collision
must have sufficient energy
35 -increasing temperature, the smaller the exponent, therefore the
larger the k
-to use the equation, either get data from two different temperature
points to determine activation energy or graphically
37 - Ea = -Rin(k2/k1) all over (1/T2 – 1/T1) = -8.314J/molK times
In(1.10x10^-5L/mols divided by 9.51x10^-9L/mols) all over (1/600K
1/500K) = 1.76 kJ/mol
38 -valley – intermediates
-peaks – steps of the reaction
-a reaction is faster because the activation energy is smaller
-the energy difference between reactants and products remains
unchanged with a catalyst
-catalyst is at the beginning and end
-intermediate only at the middle, used up, not remade
40 - how a catalytic converter works
41 -when pressure and temperature cannot be readily controlled, a
enzyme allows reactions to occur under relatively mild conditions
-do not need to know mechanism
42 -Ans1: B – compare 1 and 2 – double red, double rate – first order
-Ans2: A – compare 1 and 3 – double blue, no change in rate – 0th
order
-Ans3: B – any concentration that affects rate of reaction does not
have to be in final equation
-Compare 1 and 4
-Therefore rate = k[A][C]
EQUILIB.
2-making reactants at same rate as products
-past the dotted vertical line, chemical equilibrium achieved – same
rate
3- once at equilibrium, colour will not change because relative concentrations
will not change
5- Ans:C
6-note exponent 2 – from coefficient of NO2
-because of Arrhenius equation
7-no units
-Ans: A
8-RT to the power of (c+d-a-b) is basically RT to the power of delta n
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gas
-The part after RT is Kc
-Pa = na/V times RT = [A]RT
9-CaCO3(s) -> CaO(s) + CO2(g)
-Concentration depend on amount of CO2, not any of the solids
10 -Ans1: A
-Ans2: B
11 - equilibrium does not mean equal amounts of reactants and products
12 - Ans: D
13 -Q = P of H2 times P of Br2 divided by P of HBr squared = 0.0025
-Ans: A
14 - Ans: B
15 - equilibrium shifts to products
17 -larger volume, shift to side with more moles
-arrow points to reactants for the chemical equation
-kp does not change
18 -exothermic reaction – heat released – shift to side of products
-therefore N2O(g) plus heat
-cool down – shift to make more heat
19 -add catalyst to a reaction at equilibrium, nothing happens
-add catalyst not at equilibrium, equilibrium reached faster
20 -a) no effect
-b) no effect on eq. Position
-c) to reactant side
-Ans: D
-E) to products side
21 -set up ICE table
-I” from reactants to products: 1.000atm; 1.000atm; 0
-C” from reactants to products: -2x; -x; 2x
-E from reactants to products: 1.000atm – 2x; 1.000atm – x; 2x
-x = 0.494atm
-P of NO = 0.012 atm
-P of NO2 = 0.988atm
-Kp = (P of NO2) squared divided by (PNO) squared times P of O2 =
(0.988atm) squared over (0.012atm)squared times (0.506atm) =
1.3x10^4 UNITLESS
21 EX2.
-I from reactants to products: 2.0 M; 0 M; 0 M
-C from reactants to products: -x; +x; +x
-E from reactants to products: 2.0M – x; x; x
-Kc = [Fe3+][SCN-]/[FeSCN2+] = x squared / (2.0M – x) = 9.1 x
10^-4
-Solve for x using the quadratic equation and x = 0.042, -0.043
-The negative value of x is inadmissible
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-Therefore [FeSCN2+] = 1.958 M
-[Fe3+]=[SCN-] = 0.042 M
-can double check by plugging in the concentrations in the Kc
equation and seeing if you get 9.0x10^-4
-OR
-Assume that x is negligible compared to initial concentration
-Therefore can write x squared / 2.0 = 9.0x10^-4 therefore x = 0.043
M
EX. 3
- Kp = 1.2
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