Lecture 6 Kinetika Kesetimbangan Kimia

F₂ (g) + 2ClO₂ (g) 2FClO₂ (g)

rate = k [F₂][ClO₂]

Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the

stoichiometric coefficient of the reactant in the balanced chemical equation.

Determine the rate law and calculate the rate constant for the following reaction from the following data:

S₂O₈2- _{(aq) + 3I}- _{(aq) 2SO}

Double [I-_{], rate doubles (experiment 1 and 2)}

First-Order Reactions

rate = - [A]

t rate = k [A]

[A] is the concentration of A at any time

[A]₀ is the concentration of A at time t=0

[A] = [A]₀e-kt

ln[A] - ln[A]₀ = - kt

A  produk

k = - D[A]

[A] _Dt k = = det-1

- = k [A] - = k dt

Decomposition of N₂O₅

13.3

2N₂O₅  4NO₂ + O₂

(det) [N₂O₅] ln[N₂O₅]

0 0.91 -0.094

300 0.75 -0.29

600 0.64 -0.45

1200 0.44 -0.82

3000 0.16 -1.83

First-Order Reactions

half-life, t_½, is the time required for the concentration of a reactant to decrease to half of initial concentration (waktu yang diperlukan agar konsentrasi reaktan turun menjadi setengah

i konsentrasi awalnya)

t_½ = t when [A] = [A]₀/2

What is the half-life of N₂O₅ if it decomposes with a rate constant of 5.7 x 10-4 _s-1_?

How do you know decomposition is first order?

A product First-order reaction

half-life [A] = [A]₀/n

1

2

3

4

2

4

8

Second-Order Reactions

rate = - D[A]

Dt rate = k [A]

2 [A] is the concentration of A at any time t [A]₀ is the concentration of A at time t=0

t_½ = t when [A] = [A]₀/2 t_½ = 1

k[A]₀

Half life for second order

Zero-Order Reactions

rate = - D[A]

Dt rate = k [A]

0 _{= k}

[A] is the concentration of A at any time t

[A]₀ is the concentration of A at time t=0

t_½ = t when [A] = [A]₀/2 t_½ = [A]0

2k

[A] - [A]

₀

=

kt

Summary of the Kinetics of Zero-Order, First-Order

and Second-Order Reactions

Order Rate Law

A + B C + D

Exothermic Reaction Endothermic Reactio

activation energy (E_a) is the minimum amount of energy required to initiate a chemical tion.

si yang terbentuk sementara oleh molekul reaktansebagai akibat tumbukkan sebelu bentuk produk dinamakan kompleks teraktifasi (activated complex).

produk lebih stabil

dingkan reaktan, maka si akan diiringi dengan

asan kalor (eksotermik).

liknya, jika produk kurang l dibandingkan reaktan,

kalor akan diserap dari ungan oleh campuran yang aksi dan reaksinya bersifat

Temperature Dependence of the Rate Constant

k = A • exp -Ea/RT

E_a is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor

ln k = - -Ea R

1

T + lnA

(Arrhenius equation)

-(3) Reaction Mechanisms

The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the

reaction mechanism (tahap elementer yang mengarah pada pembentukkan produk disebut mekanisme reaksi.

2NO (g) + O₂ (g) 2NO₂ (g)

N₂O₂ is detected during the reaction!

Elementary step: NO + NO N₂O₂ Elementary step: N₂O₂ + O₂ 2NO₂ Overall reaction: 2NO + O₂ 2NO₂ +

Intermediates are species that appear in a reaction mechanism

but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step.

Rate Laws and Rate Determining Steps

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall balanced equation for the reaction.

• The rate-determining step should predict the same rate law that is determined experimentally.

The

rate-determining step

is the

slowest

step

Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

anyaknya molekul yang bereaksi dalam tahap elementer menentukan molekularitas reaksi

olecularity of reaction). Setiap tahap elementer yang baru dibahas disebut reaksi bimolecula

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.

E_a k

uncatalyzed catalyzed

rate_catalyzed > rate_uncatalyzed

Energy Diagrams

Exothermic Endothermic

(a) Activation energy (Ea) for the forward reaction

(b) Activation energy (Ea) for the reverse reaction

(c) Delta H

50 kJ/mol 300 kJ/mol

150 kJ/mol 100 kJ/mol

The experimental rate law for the reaction between NO₂ and CO to produce NO and CO₂ is rate = k[NO₂]2_{. The}

reaction is believed to occur via two steps:

Step 1: NO₂ + NO₂ NO + NO₃

Step 2: NO₃ + CO NO₂ + CO₂

What is the equation for the overall reaction?

NO₂+ CO NO + CO₂

What is the intermediate? Catalyst?

NO₃

What can you say about the relative rates of steps 1 and 2?

rate = k[NO₂]2 _{is the rate law for step 1, so} step 1 must be slower than step 2

Write the rate law for this reaction. Rate = k [HBr] [O₂]

List all intermediates in this reaction.

List all catalysts in this reaction.

HOOBr, HOBr

Ostwald Process

Hot Pt wire over NH₃ solution Pt-Rh catalysts used

in Ostwald process

4NH₃ (g) + 5O₂ (g) Pt catalyst 4NO (g) + 6H₂O (g)

2NO (g) + O₂ (g) 2NO₂ (g)

Catalytic Converters

13.6

CO + Unburned Hydrocarbons + O_{2 converter}catalytic CO₂ + H₂O

Enzyme Catalysis

Type of reactions

rous reactions

2K(s) + 2H₂O(l)  2KOH(aq) + H₂(g)

Potassium reacts with water vigorously

Formation of insoluble salts

Ag+(aq) + Cl−(aq) AgCl(s)

Very rapid reactions

Fe3+_{(aq) + 3OH}−_(aq) _Fe(OH)

ery rapid reactions

+_{(aq) + OH}−_(aq) _H

2O(l)

id-alkali neutralization reactions

pid or moderate reactions

(s) + 2Ag+_(aq)  _Zn2+_{(aq) + 2Ag(s)}

placement reactions of metals:

aq) + 2Br(aq)  2Cl(aq) + Br₂(aq)

placement reactions of halogens:

Slow reactions

Fermentation of glucose

C₆H₁₂O₆(aq)  2C₂H₅OH(aq) + 2CO₂(g)

Very slow reactions

Rusting of iron

Extremely slow reactions

CaCO₃(s) + 2H+(aq)  Ca2+(aq) + CO₂(g) + H₂O(l)

Before corrosion _{After corrosion}

occur

Amount is usually expressed in

Concentration mol dm−3

Mass g

Volume cm3 _{or dm}3

The Concept of Equilibrium

•

As a system approaches

equilibrium

, both the

forward and reverse

reactions are occurring.

•

At equilibrium

, the

forward and reverse

reactions are

A System at Equilibrium

Once equilibrium is

achieved, the amount of each reactant and

Rates become equal Concentrations become constant

A System at Equilibrium

The Equilibrium Constant

Forward reaction: _{Reverse reaction:}

The Equilibrium Constant

At equilibrium

To generalize, the reaction:

has the equilibrium expression:

K

_c

=

is depending upon the temperature (T)

a K >1 kesetimbangan akan terletak disebelah kanan, cenderung ke arah produk

Homogeneous equilibrium (kesetimbangan homogen)

Berlaku untuk reaksi yang berada pada fasa yang sama. Contoh:

K

_c

=

Konsentrasi dinyatakan dalam_{mol per liter}

suhu tetap, tekanan P dari suatu gas berbanding lurus dengan konsentrasi

m mol per liter gas tersebut; P = (n/V) RT, sehingga untuk proses kesetimbang t dituliskan sbb:

K

_p

=

Kp menyatakan bahwa konsentrasi kesetimbangan dinyatakan

dalam tekanan.

a

A

b

B

K

_c

=

K

_p

=

Dengan asumsi perilaku gas ideal, P_A V = n_A RT

P_A = n_A RT / V

Selain itu, P_B V = n_B RT

P_B = n_B RT / V

K

_p

=

nB RT / V)

n_A RT / V)

=

n_B

n_A/V dan n_B/V mempunyai satuan mol per liter dan dapat digantikan oleh [A] dan [B], sehingga

K

_p

=

(

RT

)

n ₌

K

c

(

RT

)

n

= b – a = mol produk gas – mol reaktan gas

kanan biasanya dinyatakan dalam atmosfer (atm), konstanta gas R yg digunakan alah 0,0821 L.atm/K.mol, sehingga dapat dituliskan hubungan K_p dan K_c sbb

K_p = K_c (0,0821 T)n

rite the equilibrium expression for K_C for the following reactions:

EXERCISE

(d)

(e)

ntukan persamaan untuk K_C dan K_p untuk reaksi reversible pada kesetimbangan

) 2NO(g) + O₂(g)  2NO₂(g)

Hoterogeneous equilibrium (kesetimbangan heterogen)

Reaksi reversible yang melibatkan reaktan dan produk yang fasanya berbeda menghasilkan kesetimbangan heterogen. Contoh:

Konstanta kesetimbangan:

K’

_c

=

Untuk konsentrasi suatu padatan, seperti halnya kerapatannya yaitu merupakan sifat intensif dan tidak bergantung banyaknya zat yang ada. Konsentrasi (mol/liter) dapat diubah menjadi satuan kerapatan (g/cm3_{) dan sebaliknya. Berdasarkan alasan}

tersebut, maka konstanta kesetimbangan dapat disederhanakan sbb:

K’_c = K_c = [CO₂]

mana K_c adalah konstanta baru dan dengan mudah dinyatakan dalam satu konsentras itu CO₂. Perlu diperhatikan bahwa nilai K_c tidak bergantung banyaknya CaCO3 dan CaO ng ada, sepanjang ada sedikit dari masing-masing yang berada dalam kondisi setimbangan.

s long as some CaCO₃ or CaO remain in the system, the amount of CO₂ above the solid ill remain the same.

Pada gambar (a) dan (b) kesetimbanga CO₂ akan sama pada suhu yang sama, walaupun jumlah CaCO₃ dan CaO yang ada berbeda.

PbCl

₂

(

s

)

Pb

2+

₍

_aq

_{) + 2 Cl}

−

₍

_aq

₎

ersebut pada ini.

enyelesaian:

a) Karena konsentrasi padatan tidak diketahui, maka konstanta kesetimbangan dapat ditulis dengan persamaan K_p = P_CO2 = 0,236.

b) Kita ketahui

K_p = K_c (0,0821 T)n

tika persamaan untuk suatu reaksi reversible ditulis dengan arah berlawanan, konstant

setimbangan menjadi kebalikan dari konstanta kesetimbangan asal.

K

_c

=

=

4,63 x 10-3

kan tetapi dapat juga dinyatakan dengan cara yang sama yaitu,

2NO

2

N

2

O

4

an konstanta kesetimbangannya menjadi