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Chemical Energetics: Hess’ Law & Its Applications

Written on June 29, 2011 by Sean in Chemistry Notes & Tips

Chemical Energetics is a very important topic in Thermodynamics and is a highly application topic in A-Level H2 Chemistry Examinations.

In it, Hess’ Law and the use of Energy Cycle Diagrams is of utmost importance. Let’s take a look at some of the key concepts of Hess’ Law.

Hess’ Law of Constant Heat Summation:

It is defined as the enthalpy change (ΔH) of a particular reaction is determined only by the initial and final states of the system regardless of the pathway taken.

By Hess’ Law,

Enthalpy change for Pathway 1 = Enthalpy change for Pathway 2

ΔH (for A → D) = ΔH (for A → B) + ΔH (for B → C) + ΔH (for C → D)

Uses:

Application of Hess’ Law is particularly useful when the enthalpy change of a reaction cannot be determined directly by experiment. In this cases, we will use Hess’ Law to calculate the enthalpy change from other data that is available (individual steps) and can be experimentally determined. A very common application of Hess’ Law is the Born-Haber Cycle which is used to determine the lattice energies of Ionic Compounds.

Energy Cycle Diagrams:

For calculations to be clear, we often draw an Energy Cycle Diagram which is made up of individual steps that link the initial and final states of the system.

General rules to calculate ΔH of a reaction using Hess’ Law are:

Step 1: Write down the equation representing the enthalpy change which you are required to calculate.

Step 2: Construct a fully labelled energy cycle according to the data provided, including state symbols.

Step 3: Apply Hess’ Law to calculate the required enthalpy change.

It takes some practice on questions before you can feel comfortable using it. In the next blog post, i will share with you some questions on how to use the Energy Cycle Diagram and the Hess’ Law to calculate enthalpy changes.

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Ionic Equilibrium: Question on Calculation of Solubility Product, Ksp

Written on December 16, 2010 by Sean in Chemistry Notes & Tips

(Bags of Calcium Hydroxide Powders ready for shipment. Photo credit A. Myers)

Calcium hydroxide, traditionally called slaked lime, is an inorganic compound with the chemical formula Ca(OH)₂. It exists as a colourless crystal or white powder and is obtained when calcium oxide (called lime or quicklime) is mixed, or “slaked” with water. It is of low toxicity and enjoys many applications.

Let’s check out a typical exam-based question on Solubility Product under the topic of Equilibria, that has to do with calcium hydroxide.

Question:

(a) Write an expression for the solubility product, K_sp, of calcium hydroxide, Ca(OH)₂.

(b) In a titration experiment in the school chemistry laboratory, 20.0 cm³ of an aqueous calcium hydroxide solution is completely neutralised by 18.2 cm³ of hydrochloric acid solution with a molar concentration of 0.050 moldm^-3.

(i) Determine the hydroxide ion concentration, and hence the pH of the aqueous calcium hydroxide solution.

(ii) Calculate a value of the solubility product of calcium hydroxide, indicating its units.

(c) State one use of calcium hydroxide which depends on its solubility in water.

Suggested Answers & Comments:

Try out the questions before you click the link below for suggested answers and comments:

(a) Dissociation Equation: Ca(OH)₂(aq) ↔ Ca²⁺(aq) + 2OH^-(aq)

Solubility constant, K_sp = [Ca²⁺][OH^-]²

(b)(i) The ionic equation for neutralisation is H⁺(aq) + OH^-(aq) → H₂O(l)

Mole of H⁺ = 0.050 x (18.2/1000) = 9.10 x 10^-4 mol

Since 1 mole of H⁺ reacts with 1 mole of OH^-, thus:

Mole of OH^- = 9.10 x 10^-4 mol

Therefore, [OH^-] = (9.10 x 10^-4 mol) / (20.0/1000) = 0.0455 moldm^-3

pOH = -log₁₀[OH^-] = -log₁₀(0.0455) = 1.34

pH + pOH = 14

pH = 14 – pOH = 14 – 1.34 = 12.66 ≈ 12.7 (3 s.f.)

(b)(ii) Dissociation Equation: Ca(OH)₂(aq) ↔ Ca²⁺(aq) + 2OH^-(aq)

[Ca²⁺] = 1/2[OH^-] = 0.0455/2 moldm^-3

Therefore, K_sp = (0.0455/2)(0.0455)² mol³dm^-9 = 4.71 x 10^-5 mol³dm^-9

Common Errors on Units:

Many students tend to make mistakes in determining the units of the solubility product. Simple multiplication and being more cautious will help you.

(moldm^-3) x (moldm^-3)² = (moldm^-3) x (mol²dm^-6) = mol³dm^-9

Hope you find the above discussions useful. I would love to hear from you. Leave me a comment below.

PS: If you find this blogpost useful to someone your know, feel free to send to them.

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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?