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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###### Document Summary

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 . When pressure and temperature cannot be readily controlled, a enzyme allows reactions to occur under relatively mild conditions do not need to know mechanism. 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. Making reactants at same rate as products past the dotted vertical line, chemical equilibrium achieved same rate. Once at equilibrium, colour will not change because relative concentrations will not change. Ans:c note exponent 2 from coefficient of no2 because of arrhenius equation no units. Rt to the power of (c+d-a-b) is basically rt to the power of delta n www. notesolution. com. Concentration depend on amount of co2, not any of the solids.

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