definition n factors (rate of reaction)

8/12/2019 Definition n Factors (rate of reaction)

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RATE OF REACTION

Definition

Speed at which reactants are converted intoproducts in a chemical reaction.

Fast reaction ; - time taken is short

So, the reactants is quickly converted to theproducts.

Thus, the rate of reaction is high/higher.

Slow reaction ; - time taken is long

So, the reactants is slowly converted to theproducts.

Thus, the rate of reaction is low/lower.


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What is the relation between rates of reaction withtime?

When the time taken is short, the rate of reaction is higher,When the time taken is longer, the rate of reaction is lower,

Rate of reaction is directly proportional with 1/timeRate of reaction is inversely proportional with time

Measuring the rate of reaction

Suitable changes; volume of gas liberated precipitate formation change in mass during the reaction colour changes

temperature changes pressure changes

Changes in selected quantityRate of reaction = ________________________________________

Time taken


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Average rate of reaction :The average value of the rate of reaction within a

specified period of time.

Notes:Reaction with high rate of reaction,completed in short time.

Reaction with low rate of reaction,completed in longer time.

Rate of reaction at given/ Instantaneous time :The actual rate of reaction at that instant.

(a.k.a : Instantaneous rate of reaction)

Instantaneousrate of reaction

= Gradient of the curve at that instant

Average rateof reaction

=Changes in selected quantity

____________________________________________

Time taken


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B: Factors Affecting The Rate of Reaction

Why these two graph different?

a) Total surface area of solid reactantb) Concentration of reactantc) Temperature of reactantd) Use of catalyste) Pressure of gaseous reactant


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Collision TheoryDear boys and girls to better understand ofcollision theory you must know few thingrelated to the theory which is;

Collision Effective collision Activation energy Collision frequency Effective collision frequency Energy profile diagram

* And also the chemical equation


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What is the Collision Theory?

During a reaction, the particles of the reactants must collide with each other, for bondbreaking and then bond formation to occur to

produce product .

Bond breaking : absorb heat energyBond formation : release heat energy

Those collisions which achieved a minimum activation energy and with the correctorientation will result in a reaction. These

collisions are called effective collisions .

If the particles collide with less energy thanactivation energy or with wrong orientation, itwill not result in reaction, is called ineffective

collisions.


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Effective collisions

Those collisions which achieved a minimumactivation energy

and with the correct orientation , will result ina reaction.

Activation Energy

Activation energy is the energy barrier thatmust be overcome by the colliding particles

of the reactants in order for reaction to occur


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Energy Profile Diagramsi. Exothermic Reactions

Energy

Reactants

Products

Activation Energy

Exothermic Reaction

Reaction path

Heat changes


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ii. Endothermic Reactions

Activation energy is the difference in energybetween the energy in reactants and theenergy at the peak of curve

Energy

Reactants

Products

Activation Energy

Endothermic Reaction

Reaction path


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Exothermic ReactionReactants Product

Total HeatEnergy

Higher Lower

Heat Energyduringreaction

Energyabsorbs during bondbreaking islower

Energyreleases during bondformation ishigher

Thus;Heat changes = Heat Energy in product

Heat Energy in reactant= - ve


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Endothermic ReactionReactants Product

Total HeatEnergy

Lower Higher

Heat Energyduringreaction

Energyabsorbs during bondbreaking ishigher

Energyreleases during bondformation islower

Thus;Heat changes = Heat Energy in product

Heat Energy in reactant= + ve

The conclusion is;The reaction occur when reactants collide;

a. achieved activation energyb. with correct orientation

Any Question so far?


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1 Effect of surface area/size

SMALLER size solid reactant, Bigger total surface area per volume Higher frequency of effective collision Higher rate of reaction

BIGGER size solid reactant, Smaller total surface area per volume Lower frequency of effective collision Lower rate of reaction

Sketch graph volume of gas against time


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- Graph I is more steeper than graph II- Thus, the gradient of graph I is more than

graph II- Thus, the rate of reaction for the experiment I

is higher than experimen II

Question: Why the total volume of gas issame?

Answer: the number of mole of reactantis same

HW: draw graph figure 1.6pg10


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Effect of Size When the size of fixed mass of solid reactant(name the reactant , CaCO 3) is smaller ,

The total surface area per volumeexposed to collision with other reactant(name the reactant ) particles is bigger .

Thus, the number of collision among thereacting particles at the surface of thesolid reactants increases . Frequency ofcollission is higher.

Thus, the number of collision achievedthe activation energy to become effectivecollision is also increases .

This lead to an increase in theFREQUENCY of EFFECTIVECOLLISION .

Hence, a higher rate of reaction .

Effect of ConcentrationWhen the concentration of the solution of areactant increases;


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The number of particles per unit volume of the solution of the reactant alsoincreases .

Thus, the number of collision among thereacting particles increases . Frequencyof collission is higher.


This lead to an increase in the frequencyof effective collision.


HW: draw graph figure 1.8 pg11HW: draw graph figure 1.9 pg12

Effect of Temperature

When the temperature of a reactant increases; The kinetic energy of reacting particles

will increase, so the particles movesfaster.


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Thus, the number of collision among thereacting particles increases . Frequencyof collission is higher.




HW: draw graph figure 1.10 pg13draw graph figure 1.11 pg13

Effect of Catalyst(Pg 13)Catalyst: a substance which alters the rate of

chemical reaction while it remainschemically unchanged at the end ofthe reaction.

Properties of catalyst;


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When the catalyst is presence; The catalyst allows the reactian to take

place through an alternative path whichrequires a lower activation energy .

Thus, more collision among the reacting particles are able to achive the loweractivation energy .



Energy

Reactant

Ea

Product

Ea : Activation energywithout catalyst

Ec : Activation Energywith catalyst

Ec

Reaction with

catalyst

Reaction withoutcatalyst

Reaction path


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Effect of PressureWhen the pressure of the gas of a reactantincreases;

The number of particles per unit volume of the gas of the reactant also increases .

Thus, the number of collision among thereacting particles increases . Thefrequency of collision is increase .




HW: pg. 25 EPC no. 1, 2, 4


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Sketch graph:(i) Concentration of solution against time

(ii) Concentration of solution against 1/time

So, what can you conclude from the graph?

____time

1 /s-1

Concentration of sodium thiosulfate Na 2S2O3 / mol dm

-3

Concentration of sodium thiosulfate(Na 2S2O3) / mol dm

-3

Time / s


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(iii) Temperature against time

(iv) Temperature against 1/time

So, what can you conclude from the graph?

Suhu /oC

1time _____ /s -1

Temperature / oC

time /s


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Temperature is directly proportionalto the rate of reaction

Concentration is directly proportionalto the rate of reaction

How to analyse the graph?


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First Situation:All of these experiments are using the same sizeand mass of catalyst. The temperature of thereactants remains the same.Why are there differences in the rate of reactionshown?

I and III and III I and IV [comparison]Please remember that the total volume of the gasdepends on the number of moles of the reactant.

Volume of gas / cm 3

Time / min

I II

V

t1 t2

III

IV

V/2


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Number of moles = Molarity Volume

(solution) 1000

Can you analyse the graph based from thevolume and the concetration of the reactant?


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Second Situation:All of these experiments are using the same typeof catalyst. The volume, concentration andtemperature of the reactants remains the same.Why are there differences in the early rate ofreaction shown?

The size of catalyst in exp I is smaller compareto exp II and III.

Or

I II IIIVolume of thegas / cm 3

V

t1 t2 Time / min


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The mass of catalyst in Exp I is more than exp IIand III.

OrThe size of reactant in exp I is smaller than expII and III

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Exercise:An experiment is carried out between 10 gof magnesium with 20 cm 3 hydrochloric acid

0.5 mol dm-3

. The reaction is completed in 8minutes.[Relatif atomic mass: H, 1; Cl, 35 ; Mg, 24;Molar volume of gas is 24 dm 3 at room

condition]

a) Write a balanced chemical equationfor this reaction.

b) Calculate the maximum volume of thegas liberated at room condition.


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c) Calculate the average rate of reactionwithin 8 minutes.

d) Calculate the mass of the magnesiumused in the reaction.

e) Sketch the graph of volume againsttime for these experiment

The formula: IF REACTANT IS SOLID

Number of mole = . mass .

A r or M r

IF REACTANT IS SOLUTION

Number of moles = Molarity Volume1000


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VOLUME OF THE GAS AT ROOM

CONDITION

Volume of = Number of moles 24 dm 3 the gas

Solution:a. Mg + 2HCl MgCl 2 + H 2

0.01

b. Number of moles of Mg= mass Mg

A r= 10

24

= 0.42 mol (EXCESS-BERLEBIHAN

Number of moles of hydrochloric acid


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= Concentration Volume1000

= MV1000

= 0.5 x 201000

= 0.01 mol

Mg + 2HCl MgCl 2 + H 2 0.42 ( 0.01 )

FBCE;2 mol HCl produce 1 mol H 2

0.01 mol HCl produce x 0.01 mol H 2

Thus;The no. of mole of H 2 = 0.005 mol

(1 mol of gas = 24 dm3

in room condition)

Volume of H 2 = 0.005 X 24 dm3 H2

= 0.12 dm 3 = 120 cm 3


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c. Average reaction in 8 minute

= 120 / 8 cm 3 min -1 = 15 cm 3 min -1

d.FBCE;2 mol HCl reacts with 1 mol Mg0.01 mol HCl reacts with x 0.01 mol Mg

Thus;The no. of mole of Mg = 0.005 mol

(1 mol of Mg = 24 g)

Mass of Mg = 0.005 X 24 g= 0.12 g

HW: pg. 28-29 RQ obj.Q no. 1-8 (copy-paste) pg. 29 Sub.Q no. 1 pg. 30 Ess.Q no. 2


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2. An experiment is carried out between 2 gof magnesium carbonate with 20 cm 3 hydrochloric acid 0.2 mol dm -3 .(RAM: H, 1; C, 12; O, 16; Mg, 24;Molar volume of gas is 22.4 dm 3 at s.t.p)a) Write a balanced chemical equation

for this reaction. b) Calculate the maximum volume of the

gas liberated at s.t.p.c) Draw a labbeled apparatus for

experimet.d) How to test and confirm the gas

liberated

Solution;a) MgCO 3 + 2HCl MgCl 2 + CO 2 + H 2O

b) No. mol of HCl = MV/1000

= 0.2 x 20 / 1000= 0.004 mol

FBCE;HCl CO 2


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2 mol 1mol0.004 mol 0.002 mol

No. of mol CO 2 = 0.002 mol

Volume of CO 2 = 0.002 x 22.4 dm3

= 0.0448 dm 3 = 0.0448 x 1000 cm 3 = 44.8 cm 3

d) Add 20 cm 3 lime water into a test tube, and

passed through the gas into the test tube.The lime water turns cloudy/chalky.

definition n factors (rate of reaction)

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