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An Update... but not on Chemistry Stuff

posted Dec 7, 2016, 1:05 AM by Grace Ong   [ updated Dec 17, 2016, 1:24 AM ]

The following post was first posted on Blogger on Monday, 1 July 2013.



It's been a while since I last posted. How time flies!

Between preparing for my son's DSA applications and being constantly driven up the wall by a tween and a teen taking a break from school, I have been busy being mesmerised by an assistant professor of Chinese history from the Minzu University of China (中央民族大学).

Her name is Meng Man (蒙曼), and she specialises in the history of Sui and Tang Dynasties. (I have other interests besides Chemistry, you know!) She is also the youngest guest speaker to be invited to China Central Television Network's long-running 百家讲坛 lecture series, being only 32 when she first gave her series of lectures on Empress Wu Zetian (武则天).

And boy, is she one hell of a lecturer and story-teller! And a very well-read and knowledgeable one too! I have so much to learn from her - her ability to summarise in a concise and logical sequence, her sharp political analysis and her ability to pace her lectures coherently to cover all the major events from the reign of Emperor Tang Taizong (唐太宗) to the reign of Tang Suzong (唐肃宗). Did I also mention that she is so very well-read?

She gave four series of lectures on 百家讲坛 between 2007 and 2011 — 120 episodes in total. As at today, I've only covered about 65 %. (That's because I watched some of her lectures twice!)

Here she is, in one of her lectures on the Song of Everlasting Regret (长恨歌), covering everyone's favourite topic: The Death of Yang Guifei (杨贵妃) at Mawei Courier Station (马嵬驿站):

(Watch this video on YouTube.)

I'll be back soon with my posts on Chemistry. Meanwhile, I'll continue to pick up lecturing tips and learn about Sui and Tang Dynasties from this very passionate and learned Chinese history professor.

The Curious Case of Water in Equilibrium Constant Expressions

posted Dec 7, 2016, 12:24 AM by Grace Ong   [ updated Dec 7, 2016, 12:24 AM ]

The following post was first posted on Blogger on Friday, 19 July 2013.



A student and I recently had an interesting discussion on a rather perplexing problem − should we or should we not include the concentration water in equilibrium constant expressions?

(Watch this video on YouTube.)
This is a my favourite scene from Disney's beloved classic, Alice in Wonderland.
And no, this is not the Johnny Depp version.

Why is water appearing and disappearing from equilibrium constant expressions like the Cheshire Cat,
teasing and confusing my dear students?

We were discussing Question 2(b)(iv) from the Nov 2010 Paper 3, in which students were asked to write an expression for Kc, and calculate its value for the following equilibrium by assuming that [H2O] = 55.5 mol dm−3 throughout.

GDL-gluconic acid equilibrium

As my student was taught to ignore water concentrations for reactions taking place in aqueous media, she calculated the equilibrium constant using the following expression:
GDL-gluconic acid equilibrium constant expression - without water concentration

Unfortunately for her (and for many other Singapore candidates in 2010), this was not what the Cambridge examiners were expecting. Since water was part of the reaction, examiners were expecting candidates to include the water concentration as part of the Kc expression, i.e.

GDL-gluconic acid Kc expression - with water concentration

That was why the value of the water concentration was given in the question − to be used in the calculations!

As discussed in my previous post, the concentration of water is generally omitted from the equilibrium constant expression for an aqueous equilibrium, even if water takes part in the equilibrium as a reactant or a product. However, I also cautioned that this omission should not be applied across the board without first giving due consideration to the purpose for which the equilibrium constant expression is defined.

In the case of the GDL-gluconic acid equilibrium, the question asked for the Kc expression, not an expression in which the water concentration was subsumed under a re-defined equilibrium constant. Students were thus required to include the concentrations of all reactants and products in strict accordance with the Equilibrium Law. That the value of the water concentration was given in the question should be a clue to students that this was the requirement.

Let's consider two more equilibria taken from past examination papers.

E.g. 1 Nov 2002 Paper 1 Question 8
H2O(l) ⇌ H+(aq) + OH(aq)

How should we write the Ka expression for this equilibrium?

Now, if we simply omit the concentration of water, the resulting 'Ka' expression would look exactly like the ionic product of water, Kw = [H+][OH], which can't be right!

Ka, the acid dissociation constant, measures quantitatively the extent to which acid molecules dissociate to form (H+) ions in solution. The stronger the acid, the greater the amount of H+ ions formed in relation to the amount of undissociated acid molecules. By this definition, we cannot omit the concentration of undissociated acid molecules from the Ka expression.

For a weak acid, HA, that dissociates in water as follows:

HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)
or in simplified form, HA(aq) ⇌ A(aq) + H+(aq)
acid dissociation constant expression

To write a Ka expression for H2O, we regard H2O as a weak acid HA, i.e. we can write it as HOH,

HOH(l) + H2O(l) ⇌ OH(aq) + H3O+(aq)
or in simplified form, HOH(l) ⇌ OH(aq) + H+(aq)

Thus the Ka of water should be written as follows:
acid dissociation constant expression for water

Indeed, the concentration of water was given in the question asconcentration of water in dilute aqueous solutionsmol dm−3, and students were expected to use this value (and the given equilibrium concentration of H+ ions at 25 °C) to calculate the Ka of water.

Note:
  1. Those who are sharp will notice that there IS one water concentration omitted here - that of water functioning as a base. In the definition of Ka, the concentration of water (functioning as a base) is omitted since it is in large excess as compared to the concentration of the weak acid, HA.
  2. For a detailed explanation, please refer to my earlier post which explains the relationship between Kc and Ka.
  3. The base dissociation constantKb, of water can be similarly derived by regarding H2O as a weak base B:

    B(aq) + H2O(l) ⇌ BH+(aq) + OH(aq)
    H2O(l) + H2O(l) ⇌ H2O−H+(aq) + OH(aq)
    base dissociation constant expression for water

  4. The Ka of water is exactly equal in value to the Kb of water. At 25 °C, each has a value of 1.8 × 10−16 mol dm−3, since [H2O] =concentration of water in dilute aqueous solutionsmol dm−3, and [H+(aq)] = [OH(aq)] = 10−7 mol dm−3 at this temperature.
 
E.g. 2 Nov 2005 Paper 2 Question 1(c)
D2O(l) + H2O(l) ⇌ 2HDO(l)     where D = deuterium

What is the equilibrium constant expression for this reaction?

There are two ways to approach this problem.
  • First, this is a homogeneous equilibrium, which means that the concentrations of all reactants and products must be included in the equilibrium constant expression, as dictated by the Equilibrium Law.
  • Second, comparable initial masses of D2O and H2O were used in this reaction, which meant that none of the reactant concentrations could be assumed to be approximately unchanged after equilibrium is reached. Thus we cannot re-define another equilibrium constant expression by subsuming any of the concentrations under a new constant.
Therefore, the Kc expression for this reaction is
Equilibrium constant expression for water-heavy water equilibrium

What to do with the Solids and Liquids in Equilibrium Constant Expressions

posted Dec 7, 2016, 12:01 AM by Grace Ong   [ updated Dec 7, 2016, 12:19 AM ]

The following post was first posted on Blogger on Saturday, 13 July 2013.



In a previous post, I discussed why it is not appropriate to assume the molar concentrations of solids and liquids to be one in equilibrium constant expressions.

What should we do with the solids and liquids in equilibrium constant expressions then?

Before I attempt to answer this question, let's review the Equilibrium Law:

For a general chemical reaction aA + bB ⇌ cC + dD, we can define an equilibrium constant, Kc, where

equilibrium constant expression for a general reaction

when the reaction is at equilibrium at a certain temperature.

Let's see how this law is applied to different equilibria.

Case 1: Homogeneous Equilibria

This is where all species are present in the same phase, and the application of the Equilibrium Law is straight-forward: the concentrations of all species present in the equilibrium are included in the Kc expression.

For example,
N2(g)+ 3H2(g) ⇌ 2NH3(g)
equilibrium constant expression for the reaction between nitrogen and hydrogen (Haber Process)


CH3CO2H(l) + CH3CH2OH(l) ⇌ CH3CO2CH2CH3(l) + H2O(l)
equilibrium constant for the esterification reaction between ethanoic acid and ethanol

Case 2: Heterogeneous Equilibria

This is where the species are present in different phases. These include reactions that involve solids and gases, or solids and liquids. For these cases, the concentrations of solids are omitted from their Kc expressions.

For example,
I     C(s)+ H2O(g) ⇌ CO(g) + H2(g)
equilibrium constant expression for steam-carbon reaction - without solid concentration


II    Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)
equilibrium constant (solubility product) expression for solubility equilibrium of magnesium hydroxide
 
Now, isn't this the same as assuming the concentrations of solids to be 1 mol dm−3?

It may seem so, but the answer is no. If we apply the Equilibrium Law directly to Reactions I and II, their Kc expressions should include the concentrations of the solid reactants:
I     C(s)+ H2O(g) ⇌ CO(g) + H2(g)
Kc expression for steam-water reaction - with solid concentration


II    Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)
Kc expression for the solubility equilbrium of magnesium hydroxide - with solid concentration

However these solid concentrations are meaningless in Kc expressions. As long as there is some solid present in the equilibrium, whether it is present in a small or large quantity does not affect equilibrium. We can therefore re-define a new Kc expression without the solid concentrations:

For Reaction I: C(s) + H2O(g) ⇌ CO(g) + H2(g)
equilibrium constant expression for carbon and steam reaction - without concentration of solid
where [C(s)] is a constant that can be calculated from the density of carbon used in the reaction

Similarly, for Reaction II: Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)

equilibrium constant expression for solubility equilibrium of magnesium hydroxide - without solid concentration
where [Mg(OH)2(s)] is a constant that can be calculated from the density of magnesium hydroxide

Note:
  1. The molar concentration of a solid may be calculated from its density as follows:

    how to determine concentration of a solid

  2. The Kc/ for Reaction II is known as the solubility product of magnesium hydroxide, and is represented by the symbol, Ksp.

Case 3: Aqueous Equilibria

This is where all species are dissolved in water. By convention, only concentrations of the species involved in the equilibrium are included in the Kc expression; the concentration of the solvent, i.e. water, is excluded, even if the latter takes part in the equilibrium as a reactant or a product.

For example, ethanoic acid reacts with water as follows:

III    CH3CO2H(aq) + H2O(l) ⇌ CH3CO2(aq) + H3O+(aq)

As long as the solution is dilute, the amount of water present in the solution is hardly affected by the reaction with ethanoic acid.

Because the concentration of water remains essentially the same before and after the equilibrium is reached, we can re-define the Kc expression for Reaction III as follows:
simplifying the equilibrium constant expression for the dissociation of ethanoic acid in water

Note:
  1. The concentration of pure water was calculated using the density of water (1.00 g cm−3) as follows:

    converting density of water to concentration of water

  2. The Kc/ for Reaction III is known as the acid dissociation constant for ethanoic acid, and is represented by the symbol Ka.
  3. We can similarly define the base dissociation constant, Kb, for the reaction between a weak base and water:

    B(aq) + H2O(l) ⇌ BH+ + OH(aq)
    base dissociation constant expression

  4. While the concentration of water is generally omitted from Kc expressions for aqueous equilibria involving water (either as a reactant or a product), you should not apply this omission across the board without first considering what the equilibrium constant is used for. I shall illustrate some exceptions with examples in my next post.

1 Molar Solids and Liquids in Equilibrium Constant Expressions?

posted Dec 6, 2016, 10:05 PM by Grace Ong   [ updated Dec 6, 2016, 10:06 PM ]

The following post was first posted on Blogger on Wednesday, 3 July 2013.



From time to time, I come across students telling me that they assume the concentrations of solids and liquids to be one when writing equilibrium constant (K) expressions.

One? Well, apparently, that was what these students were taught. This is really puzzling because there is no way the concentration of water in an aqueous solution or the concentration of a solid salt in a solubility equilibrium can be 1 mol dm−3.

I suspect this approach was taken because there was no distinction made between concentration and activity.

A chemical reaction depends on the activity of each species, not their concentrations. In an ideal solution, the number of ions that is available to react is exactly equal to the number of ions that is present in the solution. However reality is seldom ideal, so the number of ions that is actually available to react is less than the number of ions present in the solution. Thus there is a need for an 'effective concentration' or activity to represent the non-ideal situation.

The activity of a chemical species is a dimensionless quantity. For pure solids and liquids, it is assigned a value of one.

Using molar concentrations in K expressions (and in rate equations too) is therefore a simplification. To have a more realistic depiction of a reaction at equilibria, K expressions are sometimes written in terms of activities rather than concentrations. In such situations, solids and liquids are omitted since their activities are equal to one. This omission, however, does not automatically apply to K expressions that are written in terms of concentrations. (I shall elaborate on this in a later post.)

Note:
  1. The formal definition of activity as given in the IUPAC Gold Book is a mathematical one. The definition given on Wikipedia, while written in words, isn't that much easier to understand. If you are game enough for stuff that is way beyond the scope of the H2 Chemistry syllabus, I suggest that you get hold of the definitive text on Physical Chemistry by Peter Atkins and Julio de Paula.
  2. For purposes of 'A' Level, it suffices to assume that molar concentrations approximate activities, which is true for dilute aqueous solutions anyway.

WeWW! CSI:New York... Where There's Smoke... Death by Caesium

posted Nov 16, 2016, 10:09 PM by Grace Ong   [ updated Dec 5, 2016, 10:17 PM ]

The following post was first posted on Blogger on Monday, 22 July 2013.



Episode 2 of CSI:New York Season 9 was an interesting one - a man ate a sandwich which was laced with caesium. He died minutes later because the caesium he swallowed exploded when it came into contact with the hydrochloric acid in his stomach.

CSI New York Season 9 Episode 2 - Leonard Brooks
The face of a murderer − Leonard Brooks
Hey! It's that same guy from Criminal Minds! Uber-geek Penelope Garcia's ex!

CSI New York Season 9 Episode 2 - Murder Victim
The poor victim, after swallowing part of the sandwich

There's no question about the explosion part, given the high reactivity of caesium, as demonstrated by the following video:

(Watch this video on YouTube.)

What's puzzling is − how did the caesium survive the trip from the deli to the park, where the man ate that lethal sandwich?

Let's see...
  1. I suppose caesium was chosen because it was the most reactive, non-radioactive Group 1 metal, and would present itself in a most dramatic fashion during the reaction with stomach acid.

    semi-solid caesium in a sealed glass ampoule
    (Source: Wikimedia Commons)
    The semi-solid state of caesium, near or at room temperature
     − This photograph is part of Dnn87's element collection.
    Click here to view the rest of his beautiful photographs.

  2. The melting point of caesium is 28.4 °C. The story did not seem to take place during summer, so I presume the surrounding temperatures were low enough for the metal to be in the solid state.
  3. Apparently, it was the mayonnaise in the sandwich that was laced with caesium. Is this even practical? The simplest homemade mayonnaise contains egg yolk, lemon juice and oil (e.g. olive oil), all of which contain substances that react with highly reactive caesium (gasp!).

    Water, for instance, reacts with caesium to give off hydrogen gas. As shown in the video above, the reaction is so exothermic that the hydrogen gas produced spontaneously ignites, causing an explosion that sometimes shatters the container!

    2Cs + 2H2O → 2CsOH + H2

    Under acidic conditions, the reaction becomes even more exothermic, so the citric acid in the lemon juice and the fatty acids present in the oil are going to make things worse!


    Examples of fatty acids present in olive oil
    structure of linolenic acid
          structure of oleic acid
    linolenic acid
    (Source: Wikimedia Commons)
          oleic acid
    (Source: Wikimedia Commons)

    structure of linoleic acid
    linoleic acid

    2RCO2H + 2Cs → 2RCO2Cs + H2
    2ROH + 2Cs → 2ROCs + H2

  4. Perhaps the caesium was still coated with a protective layer of unreactive mineral oil in the sandwich?
Maybe that was the case.

No, wait! Given that the caesium was placed between bread and other fillings that contain water, fatty acids and other substances that may react with it, all that 'friction' as the sandwich was being transported from the deli to the park would have rub any mineral oil off the metal. I think the sandwich would have started 'smoking' way before the victim took a bite of it.

I think I'm thinking way too much...

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