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Requirements for GCE A-Level H2 Chemistry (9647)

Written on December 14, 2010 by Sean in Annoucement, Chemistry Notes & Tips, Exam Strategies, Tips for Improvement

In order for students to be prepared for the A-Level (Advanced Level) H2 Chemistry Exams, it is important to understand the syllabus content by topics as well as the specific learning outcomes. This will allow the student a birds-eye view of what is required for him/her to do well. Let’s take a quick look at the subject content:

Subject Content for GCE A-Level H2 Chemistry (subject code: 9647)

PHYSICAL CHEMISTRY

Atoms, Molecules & Stoichiometry
Atomic Structure
Chemical Bonding
The Gaseous State
Chemical Energetics
Electrochemistry
Equilibria
Reaction Kinetics

INORGANIC CHEMISTRY

The Periodic Table: Chemical Periodicity
Group II
Group VII
Introduction to Transition Elements

ORGANIC CHEMISTRY

Introductory Topics
Hydrocarbons
Halogen Derivatives
Hydroxy Compounds
Carbonyl Compounds
Carboxylic Acids and Derivatives
Nitrogen Compounds

In this chemistry blogsite which is meant for A-Levels H2 Chemistry (as well as H1 Chemistry), all our blog post discussions will be based on the above topical terms. As such, you can do a search for specific topic via the search bar or category located on the right hand side of this blog.

For specific learning outcomes per topic, please go to Singapore Examination and Assessment Board (SEAB) to download a copy of the syllabus requirements.

Note:

A-Levels and Integrated Programme (IP) Chemistry students taking the GCE A-Level H2 Chemistry in Singapore will be assumed to have knowledge and understanding of Chemistry at GCE O-Level (or other equivalents), as a single subject (Pure Chemistry subject code: 5072) or as part of a balanced science course i.e. Combine Science (Chemistry/Biology or Chemistry/Physics).

If you need to brush-up your GCE O-Level Chemistry knowledge, please go to my O-Level Chemistry Blog SimpleChemConcepts.com that was started 3 years ago.

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Chemical Equilibrium: Question on Equilibrium Constant, Kp

Written on September 30, 2010 by Sean in Chemistry Notes & Tips

In the previous blogpost, we have discussed on the concepts of chemical equilibrium in a reversible reaction.

We have also discussed on the formula to determine Equilibrium Constant, K_c and K_p.

For Reversible Reactions involving Gases, we should use the following formula:

Equilibrium constant, K_p = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b

Let’s apply the formula to a question that was given by my students from a local Junior College (JC).

Now, at equilibrium, reactant (X) and products (Y and Z) will be all present together.

Since the question stated that the pressure is constant (value is P) and the temperature is also constant, we can assume that the total pressure, P_T of the gases at equIlibrium is P.

By stoichiometry ratio comparison of equation

X <=> Y + Z;

Since Mole of Y = Mole of Z at equilibrium, therefore

Partial Pressure of Y, P_Y = Partial Pressure of Z, P_Z

Since equilibrium partial pressure of X, P_X is 1/7 P, it means that the other 6/7 P of pressure must be shared equally among Y and Z.

Thus,

Partial pressure of Y, P_Y = 3/7 P

Partial pressure of Z, P_Z = 3/7 P

Using the formula on Equilibrium Constant, we will have:

K_p = (P_Y)(P_Z) / (P_X) = (3/7 P)(3/7 P) / (1/7 P) = 9/7 P

Answer: B

Are you learning something?

Hope you do!

If you think the above information is useful to your friends, classmates or siblings, feel free to share it with them.

PS: Do leave me your thoughts in the comments section below. I would love to hear from you.

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Chemical Equilibrium: Reversible Reaction Introduction

Written on September 30, 2010 by Sean in Chemistry Notes & Tips

A reversible reaction is one in which can proceed in both directions i.e. forward and reverse reactions.

The reactants are not completely converted to products. Instead, an intermediate position or EQUILIBRIUM is reached whereby both reactants and products are present.

This also means that reversible reactions are never complete, a mixture of reactants and products is obtained.

At equilibrium in a reversible reaction:

1. Rate of forward reaction = Rate of reverse reaction

2. Concentration of the reactants and the products are unchanged

It is important to know what is the EQUILIBRIUM CONSTANT, which is defined as the measure of the extent to which the reaction are coverted into products before equilibrium is reached.

Consider the equilibrium, aA + bB <=> cC + dD

Equilibrium constant, K_c = [C]^c[D]^d / [A]^a[B]^b and is a constant at a given temperature

For reactions involving gases, the equilibrium constant can be represented in terms of the partial pressure, P, of the gases and is give by the symbol, K_p.

Consider the equilibrium, aA (g) + bB (g) <=> cC (g) + dD (g)

Equilibrium constant, K_p = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b and is a constant at a given temperature

Something to note for Partial Pressure:

Partial pressure, P, of a gas is equal to the product of its mole fraction and the total pressure, P_T.

Example, in a mixture of two gases A and B (where n_x = no. of moles of gas X)

Partial pressure of Gas A, P_A = [ n_A / (n_A + n_B) ] x P_T

Important Note:

1. Concentration of a SOLID is constant, and is omitted in expressions of K_c and K_p.

2. State symbols are not essential in writing the expressions for equilibrium constant.

So far so good?

If you think the above information is useful to your friends, classmates or siblings, feel free to share it with them.

PS: Do leave me your thoughts in the comments section below. I would love to hear from you.

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Sep 10

Chemical Equilibrium – Question on Chemical Constant, Kp

In the previous blogpost, we have discussed on the concepts of chemical equilibrium in a reversible reaction.

We have also discussed on the formula to determine Equilibrium Constant, K_c and K_p.

For Reversible Reactions involving Gases, we should use the following formula:

Equilibrium constant, K_p = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b

Let’s apply the formula to a question that was given by my students from a local Junior College (JC).

Now, at equilibrium, reactant (X) and products (Y and Z) will be all present together.

Since the question stated that the pressure is constant (value is P) and the temperature is also constant, we can assume that the total pressure, P_T of the gases at equIlibrium is P.

By stoichiometry ratio comparison of equation X <=> Y + Z;

Mole of Y = Mole of Z

Partial Pressure of Y, P_Y = Partial Pressure of Z, P_Z

Since equilibrium partial pressure of X, P_X is 1/7 P, it means that the other 6/7 P of pressure must be shared equally among Y and Z.

Thus,

Partial pressure of Y, P_Y = 3/7 P

Partial pressure of Z, P_Z = 3/7 P

Using the formula on Equilibrium Constant, we will have:

K_p = (P_Y)(P_Z) / (P_X) = (3/7 P)(3/7 P) / (1/7 P) = 9/7 P

Are you learning something?

Hope you do!

If you think the above information is useful to your friends, classmates or siblings, feel free to share it with them.

PS: Do leave me your thoughts in the comments section below. I would love to hear from you.

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Ideal Gaw Law: Challenging Question

Written on August 26, 2010 by Sean in Chemistry Coaching, Chemistry Notes & Tips

Just like the question we have discussed in the previous blogpost, this one was sent to me by a H2 Chemistry student for discussion.

Apparently, this question was also mentioned in the A-Level Chemistry Challenging Drill Questions for H1/H2 that is available in major Singapore bookstores.

Let’s take a look at the question:

Question:

Which gas can be most easily liquefied by cooling and applying pressure?

A. Ar

B. H2

C. HF

D. CH4

Now, try it out by applying the concepts you have learned in the previous blogpost.

Check out the answer and suggested solutions below:

Answer:

Common mistakes by students & their arguments:

When i post this question to my JC1 H2 Chemistry Small-Group Class, many gave A (Argon) as the answer or B (H₂).

They argued that for gas molecules to be able to come together to become liquid, they must be very small (in size) as well as have weak intermolecular forces between their molecules. This will allow the molecules to come closer together much more easily.

It seemed a valuable argument at first thought, but after careful considerations the answer should be C (HF) if we apply the essential concepts we learned in earlier post.

With such disagreement and knowing that a book with suggested answer is available, i went to purchase a copy of A-Level Chemistry Challenging Drill Questions for H1/H2 as well as its Solutions Book.

From the book, I see the following suggested answer for the above question:

“To be easily liquefied at low temperatures and high pressures, the gas molecules must not occupy significant volumes and have strong attractive forces forces between molecules. Therefore, Ar is most easily liquefied as the attractive forces between gas molecules are the weakest.”

The answer given by the book is exactly the way some of my students think. However, this is the wrong concept! BEWARE especially if you have a copy of it.

Ideal Gas Law: Exam-based Question

Written on August 23, 2010 by Sean in Chemistry Notes & Tips

The following is a question adapted from a JC’s H2 Chemistry Test that was emailed to me by a JC student.

Let’s use the essential concepts that we have learned in the previous blogpost to solve this MCQ question.

Question:

Which gas is most likely to deviate most from ideal gas behaviour?

A. Ar

B. N₂

C. CH₄

D. HCl