Secondary 4 Chemistry Notes & Resources
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As the carbonyl group is polar, both aldehydes and ketones are polar simple covalent molecules. However, they are unable to form intermolecular hydrogen bonds as they do not have a H atom bonded to a F/O/N atom. The table below highlights some of the important physical properties of carbonyl compounds:
|(i) Both aldehydes and ketones have higher boiling points than alkanes with similar number of electrons|
(ii) Aldehydes and ketones have lower boiling points than alcohols or carboxylic acids with similar number of electrons
(i) Soluble in water (smaller aldehydes and ketones)
(ii) As the hydrocarbon length increases, solubilities of aldehydes and ketones decreases
|In organic solvent|
(i) Generally soluble
The most common chemical reaction undertaken by carbonyl compounds is the nucleophilic addition reactions. A typical nucleophilic reaction is the addition of the HCN molecule to the carbonyl functional groups.1. Why do carbonyl compounds attract nucleophiles?
Answer : The C atom of the -C=O group (the carbonyl carbon) bears the partial positive charge as it is bonded to a more electronegative oxygen atom. As a result, electron rich nucleophiles are attracted to this electron-deficient site. 2. Why do carbonyl compounds undergo addition reactions?
Answer : Simply, there is a C=O bond that is unsaturated.
Now, let’s take a look at the Nucleophilic Addition mechanism below:
REAGENTS and CONDITIONS:
HCN + trace NaOH (aq) / trace NaCN (aq), cold
Note: the role of HCN in the reaction is a bronsted acid while the role of NaCN in the reaction is to act as a catalyst by providing CN⁻ nucleophile initially.
Checklist when drawing nucleophilic addition mechanism
- Name of the mechanism (nucleophilic addition)
- Steps in sequential order (generation of nucleophile, nucleophilic attack on carbonyl carbon, regeneration of nucleophile)
- Curved arrows showing movement of electrons (from nucleophile to carbonyl C, from C=O bond to O, from O⁻ to H in HCN, from H-C bond to C in HCN)
- Charges and lone pair on intermediates and nucleophiles
- Labels (fast, slow)
Just like in SN1 mechanism, the compound attacked by the nucleophile is trigonal planar in shape. Hence, in nucleophilic addition, the CN – nucleophile has an equal chance of attacking the trigonal planar C=O group from above or below the plane. If the resulting molecule is chiral, a racemic mixture is produced and the product is optically inactive.
Between aldehydes and ketones, aldehydes are generally more reactive than ketones to nucleophilic attack. The table below highlights two main reasons.
|Steric Factor||Electronic Factor|
|Bulky hydrocarbon groups increase steric hindrance about the carbonyl carbon, hindering the approach of the attacking nucleophile hence ketones are less reactive than aldehydes as they are less susceptible to attack by nucleophiles||Electron donating alkyl or aryl groups reduce the partial positive charge (electron deficiency) on the carbonyl carbon and hence decreases the susceptibility of the carbonyl carbon to nucleophilic attack, making ketones less reactive than aldehydes|
As mentioned in previous chapters of organic chemistry, distinguishing tests make up an important bulk of concepts. In the chapter of carbonyl compounds, students are introduced to many different types of distinguishing tests to identify various compounds. Hence, the table below would provide a concise summary of the different tests that students are expected to familiarise with.
|To distinguish||Reagent and conditions||Observation|
|Carbonyl compounds from other functional groups||2,4-DNPH, warm||Carbonyl compounds: orange ppt formed Non-carbonyl compounds: no orange ppt formed|
|Aldehydes from ketones||KMnO4 (aq), H2SO4 (aq), heat||Aldehydes: Purple KMnO4 is decolourised Ketones: Purple solution remains purple|
|K2Cr2O7 (aq), H2SO4 (aq),heat||Aldehydes: Orange K2Cr2O7 turns green Ketones: Orange solution remains orange|
|Tollens’ reagent, warm||Aldehydes: silver mirror is formed Ketones: No silver mirror formed|
|Aliphatic aldehydes from aromatic aldehydes||Fehling’s solution, warm||Aliphatic aldehydes: reddish brown ppt is formed Aromatic aldehydes: no reddish brown ppt seen|
|Methyl carbonyl compounds (and methyl alcohol compounds as mentioned previously in hydroxy compounds chapter)||I₂ (aq), NaOH (aq), warm||Methyl carbonyls (and methyl alcohols) : yellow ppt is formed Others: No yellow ppt observed|
Test Yourself (Part I):
Suggest a structure for each of the isomers, A,B and C of the compound C3H6O2 , based on the following reactions, Explain which functional groups in each molecule are taking part in the reaction.
- A gives tri-iodomethane and reduces Fehling’s solution
- B gives tri-iodomethane but does not reduce Fehling’s solution
- C does not give tri-iodomethane but does reduce Fehling’s solution
1. A gives tri-iodomethane and reduces Fehling’s solution
The methyl group gives tri-iodomethane while the aldehyde group reduces Fehling’s solution.
2. B gives tri-iodomethane but does not reduce Fehling’s solution
The methyl group gives tri-iodomethane while the ketone group is unable to reduce Fehling’s solution.
3. C does not give tri-iodomethane but does reduce Fehling’s solution
There is no methyl group to give tri-iodomethane and the ketone group is unable to reduce Fehling’s solution.
Test Yourself (Part II):
Quadratic acid, used medically in the treatment of wart, is an unusual organic compound with molecular formula, C4H2O4 and has the following structure:
- Suggest what you would see when quadratic acid reacts with 2,4 dinitrophenylhydrazine. Write a balanced equation for the reaction.
- Describe a test to distinguish between quadratic acid and CHOCH2CH2CHO.
1. Orange precipitate
2. Reagents and Conditions: Tollens’ reagent, warmObservations:
Quadratic acid: No silver mirror
CHOCH2CH2CHO: Silver mirror formed
Organic chemistry questions most notoriously come in the form of elucidation questions. But not to fear, they are conquerable.
Let’s walk through a sample elucidation question
Worked Example 1
|Observation||Type of reaction||Functional Group present|
|Reacts with NaBH4 to produce C, which consists of 50:50 mixture of 2 isomers||Reduction||Diol present in C Chiral carbon atom present in C|
|Aldehydes from ketones||KMnO4 (aq), H2SO4 (aq), heat||Aldehydes: Purple KMnO4 is decolourised Ketones: Purple solution remains purple|
|Reacts with Al2O3||Elimination||Carbon carbon double bond present in D|
|No reaction with 2,4-DNPH||Absence of condensation||Absence of carbonyl functional group in D|
|Aqueous bromine decolourised||Electrophilic addition||Carbon carbon double bond present in D|
Tip: Always list out the observations (stated in the question) and deductions (type of reaction, functional group present, etc) from the information provided neatly in a table form.
Introduction to Organic Chemistry
Organic chemistry makes up a huge portion of the H2 chemistry syllabus and mainly deals with the structure, properties and reactions of carbon-containing compounds.
Some important definitions that students should definitely memorise are
A Empirical formula
B Molecular formula
D Free radical
F Homolytic bond fission
H Heterolytic bond fission
|Other substituents and/or functional groups||Shows the number of C in the|
|Principal functional group|
|Example : 2-chloropentanoic acid = CH₃CH₂CH₂CH(Cl)COOH|
|"2-chloro" is the prefix||"pentan" is the stem||"oic acid" is the suffix|
Worked Example 1
A , CH₃CH₂CH(CH₃)C(CH₃)₂CH(NH₂)CH(Cl)COOH
|Step 1 - Suffix and root|
(i) Identify the principal functional group
(ii) Select the longest continuous C chain containing the principal functional group and count the number of C atoms
|(i) principal functional group : -COOH|
Suffix : -oic acid
(ii) number of C in parent chain : 7
Root : heptan-
(all single bonds in the carbon chain)
|Step 2 - Prefixes|
Identify all the remaining functional groups,
showing them as the prefixes
|Other functional groups : CH₃ , NH₂ , Cl|
Prefixes : methyl, amino and chloro
|Step 3 - Prefixes|
Number the C atoms in the main chain from
one end such that
(i) the lowest number of given to the group
cited as the suffix
(ii) the lowest possible individual numbers
to the groups cited as the prefixes
|(i) position of principal functional group: on C1|
(ii) position of the amino functional group: on C 3
(iii) position of the chloroalkane functional
group: on C 2
(iv) position of the 3 methyl substituents :
two on C 4 and one on C 5
|Step 4 - String prefixes together|
- If the same substituent appears more than once, indicate using di- , tri- , tetra- etc.
- The appropriate positional number is still required even if the groups are connected to the same carbon on the main chain
- Different prefixes are listed in alphabetical order (multiplying prefixes like di- tri- etc does not affect the alphabetical order)
- Commas (,) are used to separate numbers while hyphens (-) are used to separate numbers and words
|Step 5 - String prefixes , root and suffix together||Hence IUPAC name for compound A is|
A huge portion of this chapter covers the concept of Isomerism:
|Chain isomers : different parent carbon chain length||Pentane||2-methylbutane|
|Positional isomers : same functional groups but at different positions||1-chloropentane||2-chloropentane|
|Functional group isomers : Different functional groups||Butanoic Acid||Methyl Propanoate|
Cis-trans Isomerism arises due to:
● Restriction to free rotation of certain parts of the molecule caused by multiple bonds, ring structure, steric factors
● Occurs in alkenes that do not have identical group bonded to each of the 2 C atoms of the C = C bond Cis-isomer: 2 same groups on same side of the bond
Trans-isomer: 2 different groups on opposite side of the bond
- Molecules must have at least 1 chiral C atom (C atom bonded to 4 different substituents)
- Molecules have no internal plane of symmetry
- Have identical physical properties
- Distinguished by their ability to rotate plane-polarised light in the opposite direction by the same extent/degree
The structure of penta-1,4-diene is shown below.
- Does penta-1,4-diene exhibit cis-trans isomerism? Explain your answer.
- Penta-1,4-diene has many structural isomers. Draw the structure of one structural isomer which exhibits sp hybridisation.
- Tip: Think of the characteristics of cis-trans isomerism.
- No. In order for cis-trans isomerism to exist, each of the two double-bonded carbon atoms in a C=C double bond must be attached to two different groups, which is not true in this case.
- Firstly, think of what sp hybridisation is. It occurs when a C atom has to form 2 σ bonds and 2 π bonds, and sp hybrid orbitals are linearly arranged to form 180° angle.
This is an extremely common section in which most students have problems understanding the concept of hybridisation and struggle to tackle related questions.
Hybridisation involves combinations of s and different numbers of p orbitals to yield degenerate (same energy level) hybrid orbitals.
- sp3 hybridisation
- sp2 hybridisation
- sp hybridisation
Tip: Hybrid orbitals are always used to form σ bonds and unhybridised 2p orbitals are used to form π bonds
1. sp3 hybridisation
● Occurs when C has to form 4 σ bonds
● sp3 hybrid orbitals arranged in tetrahedral structure with bond angle 109.5°
2. sp2 hybridisation
● occurs when C has to form 3 σ bonds and 1 π bond
● sp2 hybrid orbitals arranged in trigonal planar shape with bond angle 120°-
● remaining unhybridised 2pz orbital lies perpendicular to the sp2 plane
3. sp hybridisation
● occurs when C has to form 2 σ bonds and 2 π bonds
● sp hybrid orbitals arranged linearly with bond angle 180°
● remaining 2 unhybridised 2p orbitals lie perpendicularly to each other
The structure of penta-1,4-diene is shown below.
- Label clearly the types of hybridisation present in the diagram.
- The bonding in penta-1,4-diene can be described as a mixture of σ and π bonding. Briefly explain, by using an appropriate diagram, the term "π bonding".
- Recall that there are only 3 types of hybridisation: sp3, sp2 and sp. Consider the hybridisation of C the 5 C atoms; observe and count the number of σ and π bonds around each C atom.
- As shown in the diagram drawn below, π bonding results from side-on overlap between two p-orbitals. (Recall Chemical Bonding)
Carboxylic Acids & Derivatives
Firstly, let's take a look at some of the important physical
properties of both carboxylic acids and acyl chlorides.
|Carboxylic acid||(i) Carboxylic acids have higher boiling points than its corresponding alkanes|
(ii) Carboxylic acids have higher boiling points than its corresponding alcohol
(iii) Boiling point increases with increasing length of alkyl chain of the carboxylic acid
|(i) The first four members of the aliphatic acids are completely miscible in water|
(ii) As the length of the non polar hydrocarbon chain increases, solubility in water decreases
|Acyl chlorides and esters||(i) Acyl chlorides and esters have lower boiling points a than the parents carboxylic acid||(i) For both acyl chlorides and esters, as size of the hydrocarbon chain increases, solubility in water decreases|
(ii) Esters are generally soluble in non-polar solvents
(iii) Acyl chlorides are soluble in non-polar solvents and soluble in water
Carboxylic acid undergoes several chemical reactions. Students should know the workings of each reaction well. The diagram below will link the reactions and products, helping students to better recall the chemical reactions undertaken by carboxylic acids.
Carboxylic acids are weak acids due to partial dissociation in water. However, they are more acidic than alcohols and phenols.
There are mainly 3 factors affecting the acid strength of a carboxylic acid:
|Nature of R group||Number of substituents||Position of substituents|
Cl-CH2COOH is a stronger acid than CH3-CH2COOH due to the presence of Cl which is an electron withdrawing group
CCl3COOH is the most acidic, followed by CHCl2COOH, then CH2ClCOOH
CH3-CH2-CHCl-COOH is a stronger acid than CH2Cl-CH2-CH2-COOH as the electron withdrawing Cl is closer to the carboxylate ion in the former acid, dispersing the negative charge to a greater extent, causing CH3-CH2-CHCl-COOH to be a stronger acid
Let's look at a worked example below:
Worked Example 1:
When compounds W, X and Z are added to separate portions of water, solutions are formed with pH 0.5, 2.5 and 3.0 (not exactly in that order). When aqueous silver nitrate is added to these three solutions, two show no reaction but the third one produces a thick white precipitate.
Answer: pH of W is 3.0, X is 2.5, Z is 0.5.
For Z, the presence of the electron withdrawing Cl group bonded directly to the C atom makes the C atom more electron deficient. Hence Z is the most reactive and likely to give up a proton, the C-Cl bond undergoes hydrolysis in water to form HCl, a strong acid. The white ppt of AgCl is formed when AgNO3 is added.
W and X are both R-COOH, hence are weakly acidic and have a higher pH than Z
For X, the presence of an electron withdrawing Cl group further disperses the negative charge
on the ion of X, hence stabilizing the conjugate base further. POE of X's dissociate lies further to
the right and it is more acidic than W.
Y had one more Cl atom than X. The presence of another electron withdrawing group further disperses the negative charge on Y's ion, hence stabilising it's conjugate base further. POE of Y's dissociation lies further to the right than that of X, thus it is more acidic than X.
Reactions of Alcohol, Phenol and Carboxylic Acid: (Recall this summary table)
Now, let's look at a worked example below:
Worked Example 2:
Suggest three distinguishing tests to differentiate between:
Each test must only react positively with one of the compounds.
You should also state what you would observe for each compound in each test.
Firstly, compare and note down the differences between the 3 compounds. Recall the distinguishing tests that are unique to deduce certain functional groups attached to a compound. Tip: Drawing out the structural form of A would help you to visualise better and identify the functional groups present clearly.
|Reagents & Conditions||Observations for A||Observations for B||Observations for C|
|Na2CO3 (aq)||No effervescence is observed.||No effervescence is observed.||Effervescence is observed. Colourless gas evolved forms white ppt with Ca(OH)2 (aq). C is confirmed.|
|Br2 (aq) OR neutral aqueous FeCl3||Orange solution remained. OR No violet colouration.||Orange solution decolourised. B is confirmed. OR Violet colouration is observed.||Orange solution remained. OR No violet colouration|
|I2 , NaOH (aq), warm OR K2Cr2O7 (aq) + H2SO4 (aq), heat||Yellow ppt formed. A is confirmed. OR Orange dichromate solution turned green.||No yellow ppt. OR Orange dichromate solution remained.||No yellow ppt. OR Orange dichromate solution remained.|
We have previously provided a diagram detailing the chemical reactions of carboxylic acids. Acyl chlorides similarly has various reactions it can take part in. The diagram below shows such:
Hydrolysis of Acyl Chlorides
The following is a concise summary of the hydrolysis reagents & conditions, reasons for rate of hydrolysis of the various types of organic compounds, and the different distinguishing tests to differentiate the organic compounds from each other.
Gist of Hydrolysis process: Breaking of C-Cl bond
Last but not least, esters are sweet-smelling compounds that can be formed from both carboxylic acids and acyl chlorides. The main reaction undertaken by esters is hydrolysis (be it acidic or alkaline). The diagram below will sum up the various hydrolysis reactions:
The Periodic Table
- Period 3
Before we begin, this chapter requires quite a lot of memorisation, but it is possible to memorise certain concepts through understanding and frequent application through practice questions.
These trends are commonly tested in Paper 1 MCQs and should be at your fingertips after much revision.
Test Yourself: [A Level H2 Chemistry 2013 Paper 1 A13]
For the elements in the third period of the Periodic Table, which property decreases consistently from sodium to chlorine?
- electrical conductivity
- ionisation energy
- melting point
- radius of the atom
Solution (with Detailed Explanation for good revision purposes):
Electrical conductivity increases from Na to Al due to the increase in number of delocalised electrons from 1 to 3. Si is a semiconductor/metalloid, hence it has low electrical conductivity and there is a sharp decline. As for P, S and Cl, they are simple molecules with no delocalised electrons (charge carriers) and thus the electrical conductivity graph declines to zero. There is an initial increase before the sudden decline in trend, thus Option A is wrong.
Option B: [Recall Atomic Structure - Ionisation Energy]
First ionisation energy generally increases across a period as nuclear charge increases (due to increase in number of protons) and there is a negligible increase in shielding effect (since successive electrons are added to the same valence shell), resulting in increase in effective nuclear charge and valence electrons becoming more strongly attracted to the nucleus. Hence, more energy is needed for removal of valence electrons from every atom.Since there is a general increasing trend, Option B is wrong.
Option D: [Recall Atomic Structure - Atomic Radius]
Atomic radius decreases across the period due to increase in nuclear charge (since there is increase in number of protons) and negligible increase in shielding effect (successive electrons are added to the same electron shell), resulting in increase in effective nuclear charge. Hence, valence electrons are more strongly attracted to the nucleus across the period, showing that atomic radius decreases across the period. Therefore, Option D is the right answer.
Melting point generally increases from Na to Si, followed by a sharp decrease from Si to P, and fluctuations between P to Ar. Thus Option C is false too.
When attempting MCQs, always try to give an explanation for every option (optimal when studying together and discussing with friends) to maximise and consolidate your learning for the chapter.
Summary Comparison Tables for Easier Revision:
|Na₂O (basic)||Na₂O(s) + H₂O(l) → 2NaOH(aq)|
Reacts vigorously to give alkaline solution
|MgO (basic)||MgO(s) + H₂O(l) ⇌ Mg(OH)₂(s)|
Mg(OH)₂(s) + aq ⇌ Mg²⁺(aq) + 2OH⁻(aq)
Slightly soluble to give a weakly alkaline solution
|Insoluble in water, no reaction||7|
|P₄O₁₀||P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq)|
Reacts to give weak tribasic phosphoric acid
|SO₃||SO₃(g) + H₂O(l) → H₂SO₄(aq)||About 1|
Acid-Base Behaviour of Oxides or Hydroxides
There is no need for complete memorisation of all the equations. Some of them can be derived easily, thus leaving space in your memory for other things.
- You need to recognise the acid-base nature of oxides and hydroxides of period 3. Metallic oxides and hydroxides are basic, non-metallic oxides and hydroxides are acidic. As for Al₂O₃ and Al(OH)₃, they are amphoteric in nature.
- When undergoing acid-base reaction, the products are salt and H₂O. Hence, for acid-base reactions involving HCl and basic oxides or hydroxides, those equations can be easily formed and balanced.
The following equations in the table are the ones that you must remember:
|Oxides||Reactions with NaOH|
|Al(OH)₃||Al₂O₃(s) +2NaOH(aq) +3H₂O(l) → 2Na[Al(OH)₄](aq)|
Al(OH)₃(s) + NaOH(aq) → Na[Al(OH)₄](aq)
|SiO₂||No reaction with dilute NaOH(aq)|
Reaction with concentrated NaOH at high temperature (extreme conditions):
SiO₂(s) + 2NaOH(conc) → Na₂SiO₃(aq) + H₂O(l)
|P₄O₁₀||P₄O₁₀(s) +12NaOH(aq) → 4Na₃PO₄(aq) + 6H₂O(l)|
|SO₃||SO₃(g) +2NaOH(aq) → Na₂SO₄(aq) + H₂O(l)|
Reaction of Chlorides with Water
|Chlorides||Reactions||pH of Solution|
|Na₂O||NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)|
Dissolves readily, no hydrolysis
6H2O(l) + MgCl₂(s) → [Mg(H2O)6]2+(aq) + 2Cl⁻(aq)
[Mg(H₂O)₆]²⁺(aq) ⇌ [Mg(H₂O)₅(OH)]⁺(aq) + H⁺(aq)Dissolves readily, slight hydrolysis; Mg²⁺ has a higher charge density than Na⁺, thus able to polarise, weaken
and break O-H bond in H₂O
of [Mg(H₂O)₆]²⁺(aq) to release H⁺ ions, resulting in a slightly acidic solution
AlCl₃(s) + 6H₂O(l) → [Al(H₂O)₆]³⁺(aq) + 3Cl⁻(aq)
[Al(H₂O)₆]³⁺(aq) ⇌ [Al(H₂O)₅(OH)]²⁺(aq) + H⁺(aq)
Dissolves and hydrolyse; Al³⁺ has high charge density, able to polarise, weaken and break O-H bond in H₂O
of [Al(H₂O)₆]³⁺(aq) to a greater extent to release H⁺ ions, resulting in an acidic solution
|SiCl₄||SiCl₄(l) + 4H₂O(l) → SiO₂.2H₂O(s) + 4HCl(aq)|
Acidic solution of HCl formed
|PCl₅||Case 1: Limited supply of water (cold)|
PCl₅(s) + H₂O(l) → POCl₃(s) + 2HCl(aq)
Case 2: Large amount of water
PCl₅(s) + H₂O(l) → H₃PO₄(aq) + 5HCl(aq)
Acidic solution of HCl formed
Test Yourself - Application Time:
[2017 CJC H2 Chemistry Prelim Paper 1 Qn 14]
The oxide and chloride of an element E are mixed separately with water. The two resulting solutions have the same effect on litmus.
What is element E?
Solution with Detailed Explanation:
For Option B and C, do note that both SiO₂ and Al₂O₃ are insoluble in water. The resulting solution would therefore be neutral. Both SiCl₄ and AlCl₃ will undergo hydrolysis to give acidic solutions.
PCl₅ undergoes hydrolysis to form H₃PO₄ and HCl, which causes the resulting solution to be acidic and turns blue litmus red. P₄O₁₀ would also undergo hydrolysis to give H₃PO₄, which is an acidic solution.
MgO gives Mg(OH)₂ in water which is weakly basic whereas MgCl₂ is weakly acidic.
Therefore, the answer is Option D.
Tips and tricks:
By observing trends across the period and understanding the reason behind why such trends appear, remembering them becomes an easier feat. Sometimes, blind memorisation can be tough and futile, but understanding the logic behind trends can solidify our understanding and memory.
- Notice that as we go across the period, the oxidation number of oxides increases. Why is that so?
Reason: The valence electrons available for bond formation increases.
- Following a similar train of thought, we understand that P and S show several oxidation numbers because they are able to expand their octet due to vacant and energetically accessible 3d orbitals available.
- Across period 3, the nature of period 3 oxides change from basic to acidic, and their structures from giant ionic lattice structure to simple molecular structure. This can be attributed to period 3 elements changing from metals to non-metals across the period.
- Why is Al₂O₃ amphoteric?
Reason: Al₂O₃ is an ionic oxide. Since Al³⁺ has high charge density (∝), it has high polarising power and is able to polarise the small O²⁻ ion, resulting in Al₂O₃ having some covalent character (refer to Chemical Bonding). Mixture of ionic and covalent character results in Al₂O₃ displaying both acidic and basic properties.
- Now you may ask, what about AlCl₃?
Even though Al is a metal, AlCl₃ is a simple covalent molecule.Reason: Al3+ has high charge density (∝), it has high polarising power and is able to polarise the large electron cloud of Cl- to a larger extent, compared to the smaller O²⁻ ion that has a smaller electron cloud and is less polarised.
Just as reactions vary in their speeds, they also vary in their extents. The topic of chemical equilibria will deal with how far a reaction can proceed. Students should familiarise themselves with both the qualitative and quantitative aspects of this chapter in order to secure marks in the examinations.
Definitions that students should memorise
A Reversible reaction
B A closed system
C Dynamic equilibrium
D Le Chatelier’s principle
The Equilibrium Law
This is a very important key concept in this chapter. For any reversible reaction, regardless of the composition of reactants and products at any instant, the system can be deemed using the term Qc
Difference between Kc vs Kp
Students tend to get confused and mess up these two equilibrium constants. We will break down the various differences and make it more straightforward and easily understandable.
|1. Both are equilibrium constants, and a constant at constant temperature||1. The reasons why Kc and Kp expressions do not include solids are different.|
- For Kc, concentration of a solid is always constant at constant temperature.
- For Kp, solids only add a negligible amount of pressure to the system.
|2. Both Kc and Kp expressions do not include solids.||2. For all Kp expressions, liquids are not included. However, liquids are included for some Kc expressions.|
- For Kp, liquids only add a negligible amount of pressure to the system.
- As for Kc expressions, concentration of water is not included if equilibrium is established in aqueous medium. In other words, H2O(l) is excluded if the system contains aqueous reactant, and H2O(l) is included if the system contains all (l) reactants
Construct the Kc or Kp expression for the following reactions and state its units
- Do we include every single substance in the Kc or Kp expression?
- When do we include H2O in the Kc or Kp expression?
- H2O(g) + C(s) ⇌ H2 (g) + CO(g)
- CH3COOH(l) + C2H5OH(l) ⇌ CH3CO2C2H5 (l) + H2O(l)
- 2CrO4 - (aq) + 2H + (aq) ⇌ Cr2O72- (aq) + H2O(l)
Questions involving changes in concentrations or partial pressures of reactants can quickly become confusing. But organising the information given in an initial-change-equilibrium (ICE) table helps make answering these questions more straightforward. Let’s take a look at how an ICE table can be utilised in the example below.
Worked example #1
Steam dissociates into its elements at very high temperatures. 2H2O(g) ⇌ 2H2 (g) + O2 (g)
At a very high temperature and a total pressure of 1atm, 20% of the steam dissociates into H2 and O2. Calculate the partial pressure that each of the 3 gases exert at equilibrium. Hence calculate the equilibrium constant, Kp, for the above reaction.
Let the initial amount of steam be a mol.
Answer = 5.68 × 10-3 atm
When calculating Kc or Kp of the system, only use concentrations or partial pressure respectively, never use amounts. Although in cases when the mole ratio of the product to all of the reactants is 1:1 and the volume of the container hence cancels out in calculation, omitting the volume might result in mark deduction.
Answering questions using Le Chatelier’s Principle
Marks are awarded based on certain keywords used. This checklist is handy in helping you assess the quality of your answers.
- By Le Chatelier’s principle, (state the change of condition)
- Predict which reaction is favoured, and why
- Identify how the position of equilibrium (POE) will shift
- Describe the observation (if question asks)
Worked example #2
Ammonia is an important raw material in making fertilisers and explosives. Nitrogen is obtained from the fractional distillation of liquid air and hydrogen is abstained from the steam reforming of hydrocarbons. With both nitrogen and hydrogen, ammonia can be produced using the Haber process.
N2(g) + 3H2 (g) ⇌ 2NH3(g) ΔH = -92.0 kJ mol-1
The conditions used when producing ammonia industrially as follows,
Explain the optimal conditions used. If relevant, use Le Chaterlier’s principle in your explanation.
- By Le Chatelier’s Principle, a low temperature will
- Favour the exothermic forward reaction to compensate for the lowered heat and
- POE will shift to the right
- Producing more ammonia However, at low temperatures, the rate of reaction is slow, and the system will take a long time to reach equilibrium. Hence an optimal temperature of 450℃ maximised both yield and rate.
- By Le Chatelier’s Principle, a high pressure will
- Favour the forward reaction to decrease the total amount of gases and
- POE will shift to the right
- Producing more ammonia However, extremely high pressure leads to higher cost of production and requires equipment maintenance. Hence an optimal pressure of 250 atm is used.
- A catalyst of finely divided iron increases the rate of reaction so that quilibrium yield is reached faster.
The relationship between ΔG8 and Position of Equilibrium
As we know, if ΔG can determine the spontaneity of a reaction, how does ΔG8 relate to the position of equilibrium of a reaction?
We know that,
● ΔG<0 indicates that the forward reaction is spontaneous and the forward reaction occurs at a faster rate,
● ΔG=0 indicates that the system is in dynamic equilibrium and both the forward and backward rate of reactions are occurring at the same rate,
● ΔG>0 indicates that the backward reaction is spontaneous and the backward reaction occurs at a faster rate.
From the graph above, ΔG8= Gproducts - Greactants and ΔG8 < 0 The change in Gibbs free energy, ΔG, is the derivative of the Gibbs free energy against composition of the reaction mixture graph.
|ΔGɵ >0||ΔGɵ =0||ΔGɵ < 0|
|Minimum G lies close to|
POE lies to left
Backward reaction favoured
Kc small (<1)
|Minimum G lies about equal|
reactants and products
Kc = 1
|Minimum G lies close to|
POE lies to left
Backward reaction favoured
Kc small (<1)
2020 Victoria School Sec 4 Chemistry Prelim Paper 1 Analysis
In this blog post, I’ll be sharing more on the 2020 Victoria School S4 Chemistry Prelim Paper 1 analysis.
And you’ll be surprised by the difficulty of this Chemistry paper and let’s go!
This paper is considered challenging and I would personally rate it 8/10!
Because it tests students on the deeper concepts and you’ll need to be very sure of your concepts in order to get it correct!
Download Question Paper 1 → [Paper] 2020 VS Sec 4 Chemistry Prelim Paper 1
Download Question Paper 2 → [Paper] 2020 VS Sec 4 Chemistry Prelim Paper 2
Answer Key → [Answer] 2020 Sec 4 Prelim Paper 1 & 2 Answers
Answers will be uploaded here as well in time to come after the sharing session.
Let's dive into 3 specific questions that cover the following topics in this Chemistry Prelim Paper Analysis - Chemical bonding, Acid & Bases & Mole Concept for this paper analysis.
Source: Victoria School – 2020 S4 Prelim MCQ Q11
Q11: The diagrams below show the structures of diamond and graphene, made up of carbon atoms in different arrangements.
Which of the following shows the correct properties of diamond and graphene?
Source: Victoria School – 2020 S4 Prelim MCQ Q13
Q13: Which of the following statements correctly describes two strong acids, HX and H2Y, of the same concentration?
1. H2Y has a lower pH than HX.
2. The rate of production of carbon dioxide is faster for H2Y than HX when equal volumes of both acids are reacted with sodium carbonate.
3. More hydrogen gas is produced for HX than H2Y when equal volumes of both acids are reacted with excess magnesium.
A 1 and 2 only
B 1 and 3 only
C 2 and 3 only
D 1,2 and 3
Source: Victoria School – 2020 S4 Prelim MCQ Q17
Q17: Which of the following is numerically equal to the Avogadro constant?
A number of atoms in 1 mole of hydrogen gas, H2
B number of electrons in 1 mole of helium gas, He
C number of ions in 1 mole of sodium chloride, NaCl
D number of molecules in 1 mole of oxygen gas, O2
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Secondary Chemistry Topic Revision: Energy Change
Energy Change: Exothermic vs Endothermic
Definition of enthalpy change: the amount of heat energy given out or taken in during a reaction, and is given by the equation:
∆𝐻 = 𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕
|Type of reaction||Exothermic||Endothermic|
|Definition||A chemical reaction that gives out energy (often in the form of heat) to the surrounding, resulting in a rise in temperature in surroundings.||A chemical reaction that takes in energy (often in the form of heat) from the surroundings, causing a drop in temperature in the surroundings.|
|Energy profile |
|Negative (<0)||Positive (>0)|
Common mistake #1: Assuming that energy level diagram and energy profile diagram is the same.
Energy level diagram will only show the energy levels of both the reactants and products, as well as the enthalpy change. Energy profile diagram also shows the energy levels of reactants and products, enthalpy change and the activation energy.
Common mistake #2: Using a double-headed arrow when labelling activation energy or enthalpy change.
For activation energy, the arrowhead will always be pointed upwards, whereas for enthalpy change, the arrowhead should point from reactants to products. Thus, for exothermic reactions, as the energy level of products is lower than that of reactants, the arrowhead will point downwards, while for endothermic reactions, it will point upwards instead.
Bond-breaking is an endothermic process as energy is taken in or absorbed to break a bond.
∆𝐻 = + 436 𝑘𝐽/𝑚𝑜𝑙
To form the hydrogen molecule, energy is lost or released to form the covalent bond. Specifically, 436 kJ of energy is released during the forming of 1 mole of H-H bonds. As bond-forming is an exothermic process, the enthalpy change is negative.
For a chemical reaction to occur, bonds in the reactants must be broken so that new bonds can be formed in the product.
∆𝐻 = - 436 𝑘𝐽/𝑚𝑜𝑙
Bond energy measures the strength of a chemical bond. The stronger the bond, the higher the bond energy. For hydrogen gas, the bond energy is 436 kJ/mol. This means that 436 kJ of energy is absorbed to break 1 mol of H-H bonds. Take note that for the chemical equation above, the enthalpy change is positive as energy is taken in.
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒: ∆𝐻 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑏𝑜𝑛𝑑 𝑏𝑟𝑒𝑎𝑘𝑖𝑛𝑔 (𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆) + 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑏𝑜𝑛𝑑 𝑓𝑜𝑟𝑚𝑖𝑛𝑔 (𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆)
The equation above tells us that a reaction is
- Exothermic (negative ΔH): if energy released in bond forming is more than energy absorbed for bond breaking.
- Endothermic (positive ΔH): if energy absorbed for bond breaking is more than energy released in bond forming.
Question: The energy profile diagram is that for the Haber Process.
What does the energy change E2 - E1 represent?
- The activation energy of the forward reaction
- The activation energy of the reverse reaction
- Enthalpy change of the forward reaction
- Enthalpy change of the reverse reaction
Explanations For Different Options 🙂
- The activation energy of the forward reaction will be the energy change between the reactant of the forward reaction (N2 + 3H2) and E2.
- The activation energy of the reverse reaction will be the energy change between the reactant of the reverse reaction (2NH3 or E1) and E2.
- Enthalpy change of the forward reaction will be from the reactant of the forward reaction to the product of the forward reaction (E1).
- Enthalpy change of the reverse reaction will be from the reactant of the reverse reaction (E1) to the product of the reverse reaction (N2 + 3H2).