Plotting a graph of ΔATAR/study score vs ATAR gives an interesting curve: students whose ATARs are around 50 have the most to gain from an additional study score point. Above about 90, the incremental ATAR gain from a single extra study point is probably below the margin of error given the way in which ATARs are calculated.
Tip for students: check the entry requirements for your course and make sure you meet those first. If your requires, for example, a particular score in the UMAT or in English, make sure you get that score. If your course requires a particular ATAR, make sure you get that, too. Remember that these scores are just entry requirements for undergraduate courses; not indicators of self-worth.
Recall from last week that our Periodic Table Smoothie contains the following species:
Amount present (moles)
Pressure: 718 kPa Temperature: 350 °C
Reactions of nitrogen in our 10-litre vessel
Our freshly-added 1.00 mol of nitrogen gas, N2(g), reacts with hydrogen gas to make ammonia in the following reversible (equilibrium) reaction. We will assume that the interior metal surface of the vessel is a suitable catalyst for this reaction (e.g. iron).
There are three other reactions below that might have occurred at higher temperature, but I’ve chosen not to raise the temperature of the vessel at this point. Rather, we’ll keep it at 350 °C to keep things manageable.*
*I was tempted at this point to elevate the temperature of our vessel to 500 °C so that the second reaction could take place as well. This would produce copious amounts of smelly ammonia gas, which would allow for larger quantities of interesting organic compounds to be produced later on. To keep our simulation safe and (relatively) simple, I’ve decided to keep the vessel at 350 °C. Interesting compounds organic will still form – only in smaller amounts.
The ammonia reaction above (the first equation) is actually an equilibrium reaction. That means that the reactants are never completely used up, and the yield is not 100%.
Recall from Le Châtelier’s principle that removing product from an equilibrium reaction causes the position of equilibrium to shift to the right, forming more product. This is because:
“If an equilibrium system is subjected to a change, the system will adjust itself to partially oppose the effect of the change.” – Le Châtelier’s principle
There are three reactions that will remove ammonia from our vessel while it’s being produced, and I’ve put all three of these into the simulation. One of these is the reverse of the reaction above (producing hydrogen and nitrogen gases) and the other two are described below. Let’s take a look at those other two reactions.
With what will the ammonia react in our vessel?
Ammonia can undergo the following reactions with the other things in our vessel**
**The ammonia does react with methane and beryllium as well, but only at temperatures of 1200 °C and 600 °C, respectively.
Two compounds will be formed: lithium amide and borazine. Lithium amide reacts with nothing else in the vessel, so the reaction chain stops there. Borazine, on the other hand, is much more interesting.
We’ve made borazine!
Borazine is a colourless liquid at room at temperature. It boils at 53 °C and has a structure that resembles that of benzene.
Because of the electronegativity difference of about 1.0 between the B and N atoms in the ring, borazine has a mesomer structure:
Like benzene, there is partial delocalisation of the lone pair of electrons on the nitrogen atoms.
Borazine polymerises into polyborazine!
Fascinatingly, borazine polymerises into polyborazine at temperatures above 70 °C, releasing an equal number of moles of hydrogen gas. Polyborazine isn’t particularly well-understood or well-documented, but one recent paper suggested it might play a role in the creation of potential ceramics such as boron carbonitrides. Borazine can also be used as a precursor to grow boron nitride thin films on surfaces, such as the nanomesh structure which is formed on rhodium.
Like several of the other compounds we’ve created in our Periodic Table Smoothie, polyborazine has also been proposed as a hydrogen storage medium for hydrogen cars, whereby polyborazine utilises a “single pot” process for digestion and reduction to recreate ammonia borane.
The hydrogen released during the polymerisation process will then react further with a little bit of the remaining nitrogen to produce a little more NH3(g) – but not much. Recall from earlier that the ammonia reaction is an equilibrium one, and the yield of NH3(g) at pressures under 30 atmospheres is very low. Pressure in our vessel is still only around 7 atmospheres.
Once polymerised, this would form about 12 grams of polyborazine:
As far as I’m aware, no further reactions will take place in the vessel this week.
Conclusion after adding 1.00 mole of nitrogen gas
Amount in mol
Pressure: 891 kPa (higher than before due to the addition of nitrogen gas) Temperature: 350 °C (vessel is still being maintained at constant temperature)
Next week, we’ll add a mole of oxygen gas to the vessel. Warning: it might explode.
Stock, Alfred and Erich Pohland. “Borwasserstoffe, VIII. Zur Kenntnis Des B 2 H 6 Und Des B 5 H 11”. Berichte der deutschen chemischen Gesellschaft (A and B Series) 59.9 (1926): 2210-2215. Web.
Last week, we put 1.00 mole of hydrogen gas into a cylinder. The resulting pressure was 243 kPa and the temperature was maintained steady at 20°C. This week, we’ll add 1.00 mol of helium gas, He(g), to the vessel and see what happens.
Will the helium react with the hydrogen?
No. Helium is completely inert. Hydrogen and helium will co-exist without undergoing any chemical reactions.
What will the resulting pressure be?
This is a very simple calculation. With 2.00 moles of gas in the vessel, the pressure would be double what it was before. This is known as Dalton’s law of partial pressures: the total pressure in a vessel is equal to the sum of all the pressure of the individual gases in the vessel.
Let’s convert that into pounds per square inch (psi) for easy comparison with everyday objects.
That’s about the same as a hard bicycle tyre.
How fast are the molecules moving?
Remember from last week that when our vessel contained only hydrogen gas, the molecules were moving around randomly with an average speed of 1760 metres per second.
Kinetic molecular theory states that the kinetic energy of a gas is directly proportional to the temperature of that gas. The formula for kinetic energy is shown below:
At constant temperature, heavier particles move more slowly than lighter ones. Even though they have the same kinetic energy, helium atoms at 20 °C move slower than hydrogen molecules at 20 °C because they have almost exactly double the mass. How much slower does the helium move? Let’s find out.
The molecules are moving at 1245 metres per second, or 4482 km/h. This is slower than the hydrogen gas by a factor of exactly root 2.
The molecules in our vessel could orbit the Earth in just 6 hours if they were to move in a single direction at this speed. Because the motion of particles in our gas mixture is random – they jiggle about rather than move in a single direction – they stay securely in the vessel.
Conclusion after adding helium
No chemistry’s happening in the vessel – not yet. Molecules of hydrogen and helium are simply co-existing in our vessel, bouncing off each other at different speeds and not interacting in any other way.
Hydrogen gas, H2(g): 1.00 mol
Helium gas, He(g): 1.00 mol
For some chemistry to happen, we’ll need to add the next element, lithium. We’ll do that next week.
Today, we’re going to answer the following question:
When 200 grams of ice is added to a bucket containing 1.00 litre of hot water, what’s the final temperature of the water?
To answer the question, we’re going to need to make some assumptions. We’ll take 1.000 litre of pure water at 80.00°C and add 200.0 g of ice (at -10.00°C) to it. What’s the final temperature of the water?
Part 1: Heat transfer method
The following equation can calculate the temperature at thermal equilibrium of any number of objects in thermal contact.
I love this equation because it’s several lines of maths shorter than the version taught in school. With this equation, you don’t even need to convert the temperatures into kelvin. Celsius works just fine.
Let’s set up the equation so that the addition series contains the variables in the question.
Now, let’s substitute the gives values into the equation. The specific heat capacity of water is 4200 J kg-1 K-1, and that of ice is 2100 J kg-1 K-1.
Great! Adding 200.0 g of ice to 1.000 L of water decreases the temperature from 80.00°C to 71.80°C.
But we’ve forgotten something. The ice will melt as soon as it hits the hot water. Since melting is an endothermic process, heat energy from the water will actually be absorbed, thus reducing the final temperature even further.
Part 2: Let’s take into account the fact that the ice melts!
Remember our formula from part 1.
The amount of energy required to melt ice can be calculated using the latent heat equation:
Removing that amount of heat energy from the system results in the following equation:
Great! Now, we’ve calculated that the final temperature of the water would be 57.36°C after the addition of the ice. That’s equal to 330.5 kelvin, which will be useful later.
However, we’ve forgotten to take something else into account: how much heat will be lost as radiation from the surface of the bucket?
Part 3: What’s the rate of heat loss from the bucket by radiation?
The rate of heat lost by radiation can be calculated by using the Stefan-Boltzmann equation, below.
P is the rate at which heat energy is radiated from the surface of the bucket in watts. Emissivity, e, of water is 0.95, and the surface area, A, should be around 0.0707 m2 for a one-litre bucket. Calculation of A is shown below. Assuming that the radius of the surface of the bucket is 6cm:
Plugging that value into the equation, we can find P. We’ll assume that the experiment is being conducted at room temperature and the temperature of the surroundings is 20.00°C (29.03 K).
This means that 2.928 joules of energy are emitted from the surface of the bucket every second. Ten minutes later, the bucket would have lost 1756.8 joules of energy due to radiation from the surface. But what about emission of radiation from the sides of the bucket?
Let’s say that our bucket is made from highly polished aluminium (which has emissivity 0.035) and it holds exactly 1.2 litres of water. We need to calculate the dimensions of the bucket.
Assuming it has straight sides (i.e. it’s a cylinder), the bucket had volume equal to the following formula:
The surface area of our bucket (excluding the open surface at the top) is:
The rate of energy radiation from the sides would therefore be:
It’s interesting to note how very little radiation is emitted from the shiny aluminium bucket, while lots more radiation is emitted from the surface of the water. This is because relatively ‘dark’ water has a much higher emissivity than shiny aluminium. Total emission from the bucket is therefore:
After ten minutes, the bucket would have lost the following amount of energy:
Let’s factor this amount of energy loss into our final temperature equation.
Not much energy is lost via radiation! Finally, let’s find the peak wavelength of the radiation emitted by the object using Wien’s law.
Part 4: What’s the wavelength of the radiation being emitted by the bucket?
Here’s Wien’s law from Unit 1 Physics…
The radiation emitted from the resulting bucket of water lies firmly in the infra-red part of the electromagnetic spectrum. The bucket would be clearly visible on an infra-red camera!
Next week, we’ll begin a new a Chemistry-themed project called Periodic Table Smoothie. More next week.