The following question was sent by one of my A-Level JC1 H2 Chemistry students in Singapore.
It is a very interesting MCQ question and requires the students to be very good in drawing the Energy Cycle Diagram, in order to solve the question.
She thinks that the answer should be “B”‘. Do you agree with her?
Show your working or brief explanation on arriving at your answer.
I will share with everyone on the solution soon.
“80% of the Success is about taking the Necessary Actions”
In the previous blog post, i have introduced to you the usefulness of Hess’ Law and Energy Cycle Diagram to determine the enthalpy changes of reactions that cannot be determined directly through experiments.
The two most common applications are:
A) Calculations involving Enthalpy Changes of Formation
B) Calculations involving Enthalpy Changes of Combustion
Let us look at two exam-based questions and see how we apply it to solve the questions.
A) Calculations involving Enthalpy Changes of Formation
Question 1:
Calculate the enthalpy change for the reaction CaF2 + H2SO4 → 2HF + CaSO4, given that the enthalpies change of formation of CaF2, H2SO4, HF and CaSO4 are -1220, -814, -271 and -1434 kJmol-1 respectively.
Solutions:
Energy Cycle Diagram:
By Hess’ Law, ΔHrθ = 2(-271) + (1434) – (-1220) – (-814)
= +58 kJmol-1
B) Calculations involving Enthalpy Changes of Combustion
Question 2:
Calculate the standard enthalpy change of formation of CH4 given than the standard enthalpies change of combustion of methane, graphite and hydrogen are -890.2, -393.4 and -285.7 kJmol-1 respectively.
Solutions:
Energy Cycle Diagram:
By Hess’ Law, ΔHfθ = -393.4 + 2(-285.7) – (-890.2)
= -74.6 kJmol-1
Hope the above examples are useful in your learning of this topic on Chemical Energetics.
Feel free to forward this post to anyone that you think will be useful to. Cheers!
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.
(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, Ksp, 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, Ksp = [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, Ksp = (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
(c) Calcium hydroxide is commonly used in agriculture to reduce the acidity of soils i.e. ‘liming the soil’.
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. 
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
INORGANIC CHEMISTRY
ORGANIC CHEMISTRY
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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