Advanced Chemistry Made Easy

Hi A-Level Chemistry students.

Today, we will discuss on the topic of IDEAL GAS LAW, which might be known as THE GASEOUS STATE in your Chemistry syllabus.

Now, in Singapore GCE A-Level H2 Chemistry syllabus, candidates are required to know the following:

1) Basic assumptions of the kinetic theory as applied to an ideal gas
2) Conditions necessary for a gas to approach ideal behavior
3) Limitations of Ideality at very high pressures and very low temperatures
4) General gas equation pV=nRT and the determination of M_r

In the prevIous blog posts, we have discussed on point 1, 2 and 3. You can do a search for those posts by keying in “Ideal Gas Law” on the top RHS field.

Today we will go through the general gas equation which is:

pV=nRT

which can be expressed as

pV=(m/M_r)RT since n = m / M_r

It is important to note the units to be used in each term in the equation. Examiners’ reports have always identified wrong use of units to be one of the most common error in A-Level exams.

p = pressure in Pa or N/m²
V = volume in m³
n = no. of moles of gas in mol
R = Gas Constant with a value of 8.31 JK^-1mol^-1
T = temperature in K
m = mass of gas in g

Let’s take a look at an exam-based question to see how we can apply the equation in examination.

Quick Check 1:

At 27 ^oC and a pressure of 8 x 10⁴ Pa, 1.00 dm³ of vapour was produced when 8.40 g of a compound was completely vapourised. Determine the relative molecular mass of the compound.

Suggested Solution:

Using the ideal gas equation

pV=nRT

we will have the following

pV=(m/M_r)RT

and after rearrangement we will have the following

M_r = (mRT)/pV

= 8.40 x 8.31 x (27 + 273) / (8 x 10⁴) x (1.00 x 10^-3)

= 261.8

= 262 (to 3 significant figures)

Note:

T/K = T/^oC + 273

1 dm³ = 1.00 x 10^-3 m³

Hope you have learned something today. In the next post we will look at more examples of solving questions using ideal gas equation.

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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, ΔH_r^θ = 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, ΔH_f^θ = -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!

(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

(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.

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