chemical reactions Flashcards

1
Q

Physical change

A

Physical changes (such as melting or evaporating) do not produce any new chemical substances.

These changes are often easy to reverse and mixtures produced are usually relatively easy to separate.

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2
Q

Chemical change

A

In chemical reactions, new chemical products are formed that have very different properties to the reactants.

Most chemical reactions are impossible to reverse.

Energy changes also accompany chemical changes and energy can be given out (exothermic) or taken in (endothermic).

The majority of chemical reactions are exothermic with only a small number being endothermic.

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3
Q

Effect of Concentration on the rate of the reaction

A

Explanation:

Increase in the concentration of a solution, the rate of reaction will increase

This is because there will be more reactant particles in a given volume, allowing more frequent and successful collisions per second, increasing the rate of reaction

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4
Q

Effect of Surface Area on the rate of the reaction

A

Explanation:

Increase in the surface area of the solid, the rate of reaction will increase

This is because more surface area particles will be exposed to the other reactant so there will be more frequent and successful collisions per second, increasing the rate of reaction

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5
Q

Effect of Temperature

A

Explanation:

Increase in the temperature, the rate of reaction will increase

This is because the particles will have more kinetic energy than the required activation energy, therefore there will be more frequent and successful collisions per second, increasing the rate of reaction.

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6
Q

Effect of Using a Catalyst

A

Explanation:

Catalysts reduce the activation energy as they create alternative pathways requiring lower activation energy, allowing more successful and frequent collisions.

This shows that when a catalyst is used, the rate of reaction will increase.

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7
Q

Explosive Combustion

A

Explosive combustion occurs when there are many fine particles in the air.

Many industrial processes such as metal working, coal mining or flour milling produce very fine and tiny particles.

These particles have a very large surface area and are combustible in air.

Even a small spark may cause them to ignite and since the surface area is so large, the rate of reaction can be incredibly fast, hence they are explosive.

Methane gas mixed with air in coal mines can also form an explosive mixture.

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8
Q

Interpret data obtained from experiments concerned with rate of reaction

A

Explanation:

Compared to a reaction with a reactant at a low concentration, the graph line for the same reaction but at a higher concentration has a steeper gradient at the start and becomes horizontal sooner.

This shows that with increased concentration of a solution, the rate of reaction will increase.

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9
Q

Interpret data obtained from experiments concerned with rate of reaction Particle Size

A

Explanation:

Compared to a reaction with lumps of reactant, the graph line for the same reaction but with powdered reactant has a steeper gradient at the start and becomes horizontal sooner.

This shows that with increased surface area of the solid, the rate of reaction will increase.

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10
Q

Interpret data obtained from experiments concerned with rate of reaction Catalyst

A

Explanation:

The diagram shows that when a catalyst is used, the activation energy is reduced as it creates an alternative pathway requiring lower activation energy, allowing more successful and frequent collisions.

This shows that when a catalyst is used, the rate of reaction will increase.

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11
Q

Temperature Interpret data obtained from experiments concerned with rate of reaction

A

Explanation:

Compared to a reaction at a low temperature, the graph line for the same reaction but at a higher temperature has a steeper gradient at the start and becomes horizontal sooner.

This shows that with increased temperature, the rate of reaction will increase.

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12
Q

Effect of Surface Area of a Solid on the Rate of Reaction

METHOD AND RESULT

A

Method:

Add dilute hydrochloric acid into a conical flask.

Use a capillary tube to connect this flask to a measuring cylinder upside down in a bucket of water (downwards displacement).

Add calcium carbonate chips into the conical flask and close the bung.

Measure the volume of gas produced in a fixed time using the measuring cylinder.

Repeat with different sizes of calcium carbonate chips (solid, crushed and powdered).

Result:

Smaller sizes of chips causes an increase in the surface area of the solid, so the rate of reaction will increase.

This is because more surface area of the particles will be exposed to the other reactant so there will be more frequent and successful collisions, increasing the rate of reaction.

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13
Q

Effect of Concentration of a Solution on the Rate of Reaction:

A

Method:

Measure 50 cm3 of Sodium Thiosulfate solution into a flask.

Measure 5 cm3 of dilute Hydrochloric acid into a measuring cylinder.

Draw a cross on a piece of paper and put it underneath the flask.

Add the acid into the flask and immediately start the stopwatch.

Look down at the cross from above and stop the stopwatch when the cross can no longer be seen.

Repeat using different concentrations of Sodium Thiosulfate solution (mix different volumes of sodium thiosulfate solution with water to dilute it).

Result:

With an increase in the concentration of a solution, the rate of reaction will increase.

This is because there will be more reactant particles in a given volume, allowing more frequent and successful collisions, increasing the rate of reaction.

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14
Q

Effect of Temperature on the Rate of Reaction:

INVESTIGATION

A

Method:

Dilute Hydrochloric acid is heated to a set temperature using a water bath.

Add the dilute Hydrochloric acid into a conical flask.

Add a strip of Magnesium and start the stopwatch.

Stop the time when the Magnesium fully dissolves.

Repeat at different temperatures and compare results.

Result:

With an increase in the temperature, the rate of reaction will increase.

This is because the particles will have more kinetic energy than the required activation energy, therefore more frequent and successful collisions will occur, increasing the rate of reaction.

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15
Q

Effect of a Catalyst on the Rate of Reaction:
INVESTIGATION

A

Method:

Add Hydrogen Peroxide into a conical flask.

Use a capillary tube to connect this flask to a measuring cylinder upside down in a bucket of water (downwards displacement).

Add the catalyst Manganese(IV) Oxide into the conical flask and close the bung.

Measure the volume of gas produced in a fixed time using the measuring cylinder.

Repeat experiment without the catalyst of Manganese(IV) Oxide and compare results.

Result:

Using a catalyst will increase the rate of reaction.

The catalyst will provide an alternative pathway requiring lower activation energy so more colliding particles will have the necessary activation energy to react.

This will allow more frequent and successful collisions, increasing the rate of reaction.

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16
Q

Describe and explain the effects of temperature and concentration in terms of collisions between reacting particles.

A

Particles need to have at least a minimum amount of energy to react when they collide.

This is called the activation energy.

At low temperatures only a small number of particles will have enough activation energy so the reaction will be slow.

At higher temperatures the particles have more kinetic energy so they move faster and with more energy.

The collisions are thus more energetic and there is a greater number of particles with sufficient energy to react, so the rate of reaction increases.

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17
Q

Concentration

A

Increasing the concentration means there are more particles per cm3, so there is less space between the particles.

Since there are more particles then it follows that there are more collisions, hence the rate of reaction increases.

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18
Q

When answering questions on the effect of concentration on the rate of reaction, you should mention that

A

there are more particles per unit volume (usually cm3) and this causes an increase in the rate of collisions.

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19
Q

Photochemical reactions

A

These reactions occur only when light is present.

The greater the intensity of ultraviolet light then the greater the rate of reaction.

E.g. the substitution of hydrogen atoms in methane by chlorine:

CH4 + Cl2 → CH3Cl + HCl

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20
Q

Silver salts in photography

A

Black and white photography film surfaces contain crystals of silver bromide.

When exposed to light they decompose to silver:

2AgBr → 2Ag + Br2

AgBr is colourless at low concentrations but the Ag appears grey-black.

Parts of the film appear black, grey or white depending on the exposure:

Stronger light = black or dark grey

Weaker light = light grey

Not exposed = white

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21
Q

Photosynthesis

A

This is the process in which plants produce food for reproduction and growth.

The equation is:

6CO2 + 6H2O → C6H12O6 + 6O2

The process requires sunlight and chlorophyll.

Chlorophyll is the green pigment in plants which absorbs sunlight and acts as the catalyst for photosynthesis.

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22
Q
A
23
Q

Reversible reactions

A

Some reactions go to completion, where the reactants are used up to form the product molecules and the reaction stops when all of the reactants are used up.

In reversible reactions, the product molecules can themselves react with each other or decompose and form the reactant molecules again.

It is said that the reaction can occur in both directions: the forward reaction (which forms the products) and the reverse direction (which forms the reactants).

24
Q

Chemical equations for reversible reactions

A

When writing chemical equations for reversible reactions, two arrows are used to indicate the forward and reverse reactions.

Each one is drawn with just half an arrowhead – the top one points to the right, and the bottom one points to the left.

25
Q

Example Chemical equations for reversible reactions

The reaction for the Haber Process which is the production of ammonia from hydrogen and nitrogen:

A
26
Q

Hydrated and anhydrous salts

A

Hydrated salts are salts that contain water of crystallisation which affects their molecular shape and colour.

Water of crystallisation is the water that is stoichiometrically included in the structure of some salts during the crystallisation process.

A common example is copper(II) sulfate which crystallises forming the salt copper(II) sulfate pentahydrate, CuSO4.5(H20)

Copper(II) sulfate pentahydrate is the most commonly occurring hydrate of copper(II) sulfate. Other hydrates with the formula CuSO4.5(H20) can occur depending on the reaction conditions, where the value for x can range from 0 to 5.

Water of crystallisation is indicated with a dot written in between the salt molecule and the surrounding water molecules.

Anhydrous salts are those that have lost their water of crystallisation, usually by heating, in which the salt becomes dehydrated.

27
Q

Dehydration of Hydrated Copper (II) Sulfate:

A

Hydrated Copper (II) Sulfate ⇌ Anhydrous Copper (II) Sulfate + Water

Explanation:

When anhydrous copper (II) sulfate crystals are added to water they turn blue and heat is given off, this reaction is reversible.

When Copper (II) Sulfate crystals are heated in a test tube, the blue crystals turn into a white powder and a clear, colourless liquid (water) collects at the top of the test tube.

The form of Copper (II) Sulfate in the crystals is known as Hydrated Copper (II) Sulfate because it contains water of crystallisation.

When Hydrated Copper (II) Sulfate is heated, it loses its water of crystallisation and turns into anhydrous Copper (II) Sulfate:

28
Q

Hydration of Cobalt(II) Chloride

A

When anhydrous blue cobalt(II) chloride crystals are added to water they turn pink and the reaction is reversible.

When the cobalt(II) chloride crystals are heated in a test tube, the pink crystals turn back to the blue colour again as the water of crystallisation is lost.

The form of cobalt(II) chloride in the crystals that are pink is known as hydrated cobalt (II) chloride because it contains water of crystallisation.

When hydrated cobalt(II) chloride is heated, it loses its water of crystallisation and turns into anhydrous cobalt(II) chloride:

29
Q

Both the hydration of CoCl2 and CuSO4 are chemical tests which are commonly used to detect the…

You should remember the equations and colour changes:

A

-presence of water.

-

CoCl2 + 6H2O ⇌ CoCl2.6H2O Blue to pink

CuSO4 + 5H2O ⇌ CuSO4.5H2O White to blue

30
Q

Reversible reactions and equilibrium

A

We have already seen that a reversible reaction is one that occurs in both directions.

When during the course of reaction, the rate of the forward reaction equals the rate of the reverse reaction, then the overall reaction is said to be in a state of equilibrium.

31
Q

Characteristics of a reaction at equilibrium

A

It is dynamic i.e: the molecules on the left and right of the equation are changing into each other by chemical reactions constantly and at the same rate.

The concentration of reactants and products remains constant (given there is no other change to the system such as temperature and pressure).

It only occurs in a closed system so that none of the participating chemical species are able to leave the reaction vessel.

32
Q

The reaction between H2 and N2 in the Haber process

A

When only nitrogen and hydrogen are present at the beginning of the reaction, the rate of the forward reaction is at its highest, since the concentrations of hydrogen and nitrogen are at their highest.

As the reaction proceeds, the concentrations of hydrogen and nitrogen gradually decrease, so the rate of the forward reaction will decrease.

However, the concentration of ammonia is gradually increasing and so the rate of the backward reaction will increase (ammonia will decompose to reform hydrogen and nitrogen).

Since the two reactions are interlinked and none of the gas can escape, the rate of the forward reaction and the rate of the backward reaction will eventually become equal and equilibrium is reached:

33
Q

The position of equilibrium

A

Equilibrium position refers to the relationship between the concentration of reactants and products at the equilibrium state.

When the position of equilibrium shifts to the left, it means the concentration of reactant increases.

When the position of equilibrium shifts to right, this means the concentration of product increases.

34
Q

Effect of catalyst on equilibrium position

A

The presence of a catalyst does not affect the position of equilibrium but it does increase the rate at which equilibrium is reached.

This is because the catalyst increases the rate of both the forward and backward reactions by the same amount (by providing an alternative pathway requiring lower activation energy).

As a result, the concentration of reactants and products is nevertheless the same at equilibrium as it would be without the catalyst.

35
Q

Le Chatelier’s Principle

A

Le Chatelier’s principle states that when a change is made to the conditions of a system at equilibrium, the system automatically moves to oppose the change.

The principle is used to predict changes to the position of equilibrium when there are changes in temperature, pressure or concentration.

36
Q

Le Chatelier’s Principle Effects of temperature

AND EXAMPLE

A
37
Q

Effects of pressure:

Le Chatelier’s Principle

A
38
Q

Effects of concentration:: Le Chatelier’s Principle

A
39
Q

When the conditions at equilibrium are changed, the system always responds by…

For example if the concentration is increased the system tries to reduce it by… or if the temperature is increased, thesystem will try to

A
  • doing the opposite.
  • changing the direction of the reaction
  • reduce the temperature by absorbing the extra heat.
40
Q

Oxidation and reduction

A

Oxidation and reduction take place together at the same time in the same reaction.

These are called redox reactions.

There are three definitions of oxidation. It is a reaction in which:

oxygen is added to an element or a compound.

an element, ion or compound loses electrons.

the oxidation state of an element is increased.

There are three definitions of reduction. It is a reaction in which:

oxygen is removed from an element or a compound.

an element, ion or compound gains electrons.

the oxidation state of an element is decreased.

41
Q

Oxidation state

A

The oxidation state (also called oxidation number) is a number assigned to an atom or ion in a compound which indicates the degree of oxidation (or reduction).

The oxidation state helps you to keep track of the movement of electrons in a redox process.

It is written as a +/- sign followed by a number.

E.g: O2- means that it is an atom of oxygen that has an oxidation state of -2. It is not written as O2- as this refers to the ion and its charge.

42
Q

Assigning the oxidation number

A

Oxidation number refers to a single atom or ion

The oxidation number of a compound is 0 and of an element (for example Br in Br2) is also 0.

The oxidation number of oxygen in a compound is always -2 (except in peroxide R-O-O-R, where it is -1).

For example in FeO, oxygen is -2 then Fe must have an oxidation number of +2 as the overall oxidation number for the compound must be 0.

43
Q

Ionic Equations

A

Ionic equations are used to show only the particles that actually take part in a reaction.

These equations show only the ions that change their status during a chemical process, i.e: their bonding or physical state changes.

The other ions present are not involved and are called spectator ions.

44
Q

Writing ionic equations

A

For the neutralisation reaction between hydrochloric acid and sodium hydroxide:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

If we write out all of the ions present in the equation and include the state symbols, we get:

H+(aq) + Cl– (aq)+ Na+(aq) + OH–(aq) → Na+ (aq)+ Cl–(aq) + H2O(l)

The spectator ions are thus Na+ and Cl–. Removing these from the previous equation leaves the overall net ionic equation:

H+(aq) + OH–(aq) →H2O(l)

This ionic equation is the same for all acid-base neutralisation

45
Q

Example redox equation: oxygen loss/gain

Zinc oxide + carbon → zinc + carbon monoxide

ZnO + C → Zn + CO

A

In this reaction the zinc oxide has been reduced since it has lost oxygen. The carbon atom has been oxidised since it has gained oxygen.

46
Q

Example redox equation: electron loss/gain and oxidation state

Zinc + copper sulphate → zinc sulphate + copper

Zn + CuSO4 → ZnSO4 + Cu

Writing this as an ionic equation:…

By analysing the ionic equation, it becomes clear that zinc has… equation

Copper has been reduced as its oxidation state has … and it has gained electrons:

A

Zn(s) + Cu2+(aq) + SO42-(aq) →Zn2+(aq) + SO42-(aq) + Cu(s)

  • become oxidised as its oxidation state has increased and it has lost electrons:

Zn(s) →Zn2+(aq).

decreased

-Cu2+(aq) → Cu(s)

47
Q

Use the mnemonic OIL-RIG

A

Oxidation Is Loss – Reduction Is Gain.

48
Q

Oxidising agent

A

A substance that oxidises another substance, in so doing becoming itself reduced.

Common examples include hydrogen peroxide, fluorine and chlorine.

49
Q

Reducing agent

A

A substance that reduces another substance, in so doing becoming itself oxidised.

Common examples include carbon and hydrogen.

The process of reduction is very important in the chemical industry as a means of extracting metals from their ores.

50
Q

Reducing agent

Example

CuO + H2 →Cu + H2O

expalnation

A

In the above reaction, hydrogen is reducing the CuO and is itself oxidised, so the reducing agent is therefore hydrogen.

The CuO is reduced to Cu and has oxidised the hydrogen, so the oxidising agent is therefore copper oxide.

51
Q

Identifying redox reactions

A

Redox reactions can be identified by the changes in the oxidation states when a reactant goes to a product.

52
Q

Identifying redox reactions example

Chlorine + potassium iodide → potassium chloride + iodine

Cl2 + 2KI → 2KCl + I2

expalnation

A

Chlorine has become reduced as its oxidation state has decreased from 0 to -1 on changing from the chlorine molecule to chloride ions:

Cl2(g) → 2Cl–(aq).

Iodine has been oxidised as its oxidation state has increased from -1 to 0 on changing from iodide ions to the iodine molecule:

2I+(aq) → I2(s)

53
Q

Identifying redox reactions by colour changes

A

The tests for redox reactions involve the observation of a colour change in the solution being analysed.

Two common examples are acidified potassium manganate(VII), and potassium iodide.

Potassium manganate (VII), KMnO4, is an oxidising agent which is often used to test for the presence of reducing agents.

When acidified potassium manganate (VII) is added to a reducing agent its colour changes from pink-purple to colourless.

Potassium iodide, KI, is a reducing agent which is often used to test for the presence of oxidising agents.

When added to an acidified solution of an oxidising agent such as aqueous chlorine or hydrogen peroxide, the solution turns a brown colour due to the formation of iodine.