April 23 : Equilibria

In class today, we covered the "Introduction to Equilibria" which investigated how changing conditions affects the position of equilibrium.

We conducted two experiments that demonstrated the Le Chatelier's principles taking affect onto systems of equilibrium: The Iron Thiocyanate Complex & The 'Fried Eggs' of PbI2 and KPbl3.

Both experiments showed how adding different solutions/conditions shifted the equilibrium to either the reactant side (causing a change in color) or the product side (also, causing a change in color).

Homework:
Finish the Equilibria practice questions (#1-6) we were given/ we started in class

Equilibrium 21/04/15

Equilibrium 21/04/15

EQUILIBRIUM

Material:

Covered in class included:

Concept of chemical equilibrium - is when a system has no net change in concentration of reactant and products, where forward and backward reactions are both taking place at the same time at the same rate. 

Le Chatelier - a chemist who came up with the principle that a system at equilibrium will shift so as to oppose any change made to it (counter the changes made to it).

Work in class:

To demonstrate the concept of equilibrium we were given specific amount of M&M's and were made to use them in different simulations. In each there would be one holder of M&M's who represented the reactants, and one holder of M&M's who represented the products. The reactant each time would transfer 1/2 of their quantity to the products, and the products would give 1/4 of their quantity to the reactants. Eventually the system would reach an equilibrium where both were giving each other the same amount. This was then repeated with varying amounts of M&M's and fractions given to each holder.

Mr. Carmichael then went on to demonstrate another explanation of equilibrium where colour was involved. In this demonstration we wrote on a worksheet that described each situation well, and we would explain the changes we would see. Refer to this sheet for further detail. From the experiment we learnt that if a catalyst was added to a reaction it would not effect the equilibrium position, but instead just get to the equilibrium faster.Also that if there is a higher pressure exposed to a reaction it will shift it's equilibrium to the side with less moles of gas to even it's self out.

Graphs covering equilibrium:

Homework:

The homework assigned today was to read pages 218-220 in the SL textbook, to take notes on physical equilibrium, and to highlight the sheet given at the end of class.

March 13th: Collision Theory and Rates of Reaction

In class, we learned about 6.1 Collision Theory and Rates of Reaction

The greater the probability that molecules will collide with sufficient energy and proper orientation, the higher the rate of reaction.

The key understandings for this topic are:

    · Species react as a result of collisions of sufficient energy and proper orientation

    · The rate of reaction is expressed as the change in concentration of a particular reactant/product per unit time

    · Concentration changes in a reaction can be followed indirectly by monitoring changes in mass, volume and color

    ·  Activation energy (E_a) is the minimum energy that colliding molecules need in order to have successful collisions leading to a reaction

    · By decreasing E_a, a catalyst increases the rate of a chemical reaction, without itself being permanently chemically changed.

Collision Theory:

·         For a reaction to occur, particles must collide with a certain minimum energy (the activation energy) and have the correct orientation

Factors that affect the rate of a reaction include:

    1.      Concentration

·         Solutions:

·         If you increase the concentration of a solution = more particles in the space & more collisions = faster rate of reaction

·         Changing the volume of a solution WILL NOT affect  the rate of reaction

·         Also, increasing the pressure of a gaseous reaction-

-       If you take a gas and increase its pressure (decrease volume of container) = more collisions = faster rate of reaction

    2.    Surface Area

·         Surface area also plays a role in determining how fast a reaction will occur.

·         For example: Crushing a solid up into a powder will greatly increase its surface area and hence the number of collisions possible between reactant particles. One would also expect to see an increased rate of reaction

    3.    Temperature

·         KE of particles is proportional to temperature in Kelvin

·         Activation Energy: the minimum energy colliding molecules need in order to have successful collisions leading to a reaction

·         A match is needed to start a Bunsen burner

·         Maxwell-Boltzmann Distribution

-       If you increase the temperature, particles will have more energy

-       Drop in curve of T₂ is because the area under the curve is fixed – it has to stay the same. The area under the curve is representing the total number of particles (in the sample)

-       At a higher temperature (T₂), we can see that more particles have an energy greater than or equal to the activation energy

We also did a lab that involved measuring the rate of a reaction (and observing the change in the rate of reaction when the temperature is increased or decreased):

·         When sodium thiosulphate reacts with an acid, a yellow precipitate of sulphur is formed

·         In this experiment, we measured how long it took for the sulphur to form

-       This is done by observing the reaction through a conical flask while viewing a black cross on white paper

-       The “X” is eventually obscured by the sulphur precipitate and the time noted

·         In the experiment, we investigated the effects of temperature of the reactants on the rate of reaction by warming up the solutions beforehand in a water bath

HL Solubility

The Solubility of an ionic compound is determined by the lattice enthalpy and the enthalpy of hydration. It can be shown through a Hess' Cycle

Factors that impact solubility:

Lattice Enthalpy- Increases as size decreases
Hydration Enthalpy- Increases as charge increases and size decreases

HL Energetics of Ionic Bond Formation 09/03/15

Energetics of Ionic Formation

What was covered:

The class started by handing in the previous homework (in this case, the 'Heat Death' poem).

The class then had a quick review on the HL information covered thus far in this chapter about 'Thermochemistry.' The review included a large packet covering thermochemistry and important definitions with their usage/purposes. 

Definitions:

• Atomisation - forming 1 mole of gaseous atoms
• Ionisation Energy - farming 1 mole of uni-positive ions in the gas phase
• Electron Affinity - forming 1 mole of uni-negative ions in the gas phase
• Lattice Energy - forming/ breaking 1 mole of ionic lattice from its constituent ions

Work in class:

In class the 'Born-Harper' cycle was covered step-by-step, on how to construct your own cycle.

A few examples were done in class, 2 with Mr. Carmichael and then some from the packet.

Example of 'Born-Harper Cycle':

To solve the lattice enthalpy, you would treat the cycle with Hess' law, saying enthalpy of formation is equal to the summation of all the other steps, for example:

Enthalpy of formation = (enthalpy of atomisation for Na)+(enthalpy of ionisation of Na)+(enthalpy of atomisation for Cl)+(enthalpy of electron affinity for Cl)+(lattice enthalpy)

-411=109+494+(242*.5)+(-364)+LE -> LE= -1580 kJ/mol

*It's important to note that the arrows pointing upwards will be positive values, and the arrows pointing downwards are of negative value (due to ex/endothermic reactions).

Homework:

The homework assigned for 11/03/15 was to complete the 2 last pages of questions 1-5.

Bibliography:

Born-Harper Cycle Diagram - "Further Enthalpy: Enthalpies, Born-Haber Cycle, Mean Bond Enthalpy." Further Enthalpy: Enthalpies, Born-Haber Cycle, Mean Bond Enthalpy. N.p., n.d. Web. 10 Mar. 2015.

5.3 Measuring Energy Changes

Energy is absorbed when bonds are broken, and released when bonds are formed. You can then use bond enthalpy values to calculate ∆H in a reaction. 

*Bond enthalpy values are averages taken across a range of environments

∆H= ∑ Bonds Broken - ∑ Bonds Made

Example:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

 4 C-H Bonds

 2 O=O Bonds

Which yields C(g) + 4H(g) + 4O(g)

This then forms:

2 C=O Bonds

4 O-H Bonds

So

∆H= [4(C-H) + 2(O=O)] - [2(C=O) + 4(O-H)]

Wednesday 11/2/15 - Hess' Law

In class we did further review of Hess' Law, especially practicing how to use it to solve problems. For example, we had to practice reading a problem and drawing the Hess' Cycle involving various other equations to try and solve the ∆H for the original equation. We also had to read about the laws of thermodynamics, which are useful when learning about enthalpy.

Laws of Thermodynamics:

1. Conservation of energy (energy is neither created nor destroying, merely transferred)
2. Entropy will always increase
3. Reaching absolute zero in temperature (which would also be zero mass and zero motion) is impossible

Additional Review:

∆H for reaction is the same, no matter which route it takes, as long as conditions are the same

∆Hc: ∆H for complete combustion of one mole of a substance with oxygen
∆Hf: ∆H for forming one mole of substance from elements in standard state

Standard conditions: 298K (or just fixed temp.), 1 atm (or 100 KPa), 1M solutions

Bond enthalpy: ∆H to break bonds

Today, we began to review for the quiz on 5.2 that will be this Wednesday.

To review, we went over the Heat of Combustion for Magnesium Challenge, whereas from the results of a theoretical experiment, we processed the proposed data as to see what the Heat of Combustion for Magnesium was.

With help from each other we solved the task, though answers differed.

Hess' Law

3rd February 2015

In today's class we carried out a lab to demonstrate Hess' Law and accordingly increase our understanding of said law. The aim of the lab was to determine the enthalpy change of a particular reaction. However, since the actual reaction is immeasurable, we needed to combust two separate compounds in order to determine the  total enthalpy change of the reaction.



Hess' Law states that the route taken doesn't impact the final enthalpy change between reactants and products. Therefore, utilizing two separate reactions, we were able to find the enthalpy change of a single reaction.

We also continued to work on our lab reports in class. The aim of the lab was to determine the relationship between the enthalpy change of combustion of alcohols and their carbon chain length.

Exothermic, Endothermic & Calorimety (23/01/15)

Exothermic

• A reaction is exothermic when it releases heat into the surrounding environment
• 
• The thermal energy of the reactants are higher than the thermal energy of the products
• The reaction as well as it's surrounding will often gain heat
• Examples: Burning, Neutralization reactions
• 

Endothermic:

• A reaction is endothermic when it absorbs heat energy from its surrounding enviroment
• 
• The thermal energy of the reactants is lower than the thermal energy of the products
• The reaction as well as it's products get colder
• Examples:Electrolysis, Reaction between ethanoic acid and sodium carbonate
• 

During the lesson today we carried out a series of experiments to determine whether a reaction was exothermic or endothermic.

To do this we had to measure the temperature difference of the substance before the reaction to the substance after the reaction.

We carried out the experiments in a styrofoam cup to deuce the amount of heat energy that was lost to the surrounding environment.

We used the heat capacity of water and the temperatures we recorded to determine how much energy was lost or gained by the system.

20 January

20 January 2015
Bea

Today's class consisted more of lab work.

The aim of the lab was to find out if solutions of different oxides were acidic or alkaline.
The four oxides we tested were sulphur trioxide, ferric oxide, magnesium oxide, and carbon dioxide.

In order to make the oxides, we had to react substances with oxygen. For some of the substances a higher concentration of oxygen was needed rather than the one found naturally in the air.
After heating the substance using the bunsen burner, the substance was reacted with oxygen in a test tube. After adding water to the test tube, we had a solution of oxides.

To test if the solutions were acidic or alkaline, we added universal indicator to the solutions.
We were able to decipher the acidity of the solutions with varying degrees of success.

Generally a rule could be applied to the oxides.

Non-metals make acidic oxides
Soluble metals make alkaline oxides.

The iron used in the reaction did not make a soluble oxide and therefore it was difficult to obtain results from the experiment.

After the experimental part, we answered questions that related to the experiment including a question that related back to acid rain.

January 14th

Today in class we did an experiment about Halogens. Halogens are the non-metals in the 17th group of the period table. They consist of Fluorine, Chlorine, Bromide, Iodine, and Astatine. They are very reactive because they have 7 valence electrons, therefore to have a full octet they only need one electron. Reactiveness decreases down a group, therefore Fluorine in the most reactive element in group 17. This is because as the atoms go down a group more electrons are added to the element so the atomic radius increases. So the pull from the protons is weaker, therefore they react less.

During our experiment we were reacting Halogens with other Halogens that were bonded with Potassium. We found that the more reactive Halogens replaced the less reactive Halogens in the potassium solution.  In our experiment we tested this by reacting Iodine, Chlorine, or Bromine Potassium solution with Iodine, Chlorine or Bromine. If there was a reaction we could deduce that the stronger Halogen had replaced the weaker Halogen in the potassium solution. For example, 

2KBr (aq) + Cl2  = 2 KCl (aq) + Br2 

We can see there that the Chlorine oxides or displaced the Bromine because it was more reactive. 

January 9th 2015

Pardon (Again) for the late blog.

On the first lesson of the semester we were introduced to our latest topic known as periodicity(i.e. trends in chemical and physical properties across the periodic table).

Trends reviewed included shielding, overall charge, metals and non-metals, metallic character, electro-negativity, electron affinity, and other similar molecular properties covered last semester.

The homework given was to fill out a periodic table with the aforementioned trends, in order to review these trends.

January 12th: Periodicity

At the start of class, we reviewed some of the periodic table trends we had studied last class- shielding, effective nuclear charge, electronegativity, electron affinity, ionization energy, and atomic/ionic radius.



Then, in groups, we looked at some of these trends more closely- discussing what they mean, their pattern, and the causes for these patterns. Specifically we mentioned attraction, distance,  shielding, effective nuclear charge, and stability of the configuration. We looked at the impact, if any, of these factors pertaining to ionization energy trends, atomic/ionic size trends, and electronegativity/electron affinity trends.

The next thing we did was observe reactions of lithium and sodium in water as an example for the reactivity of alkali metals. We then talked about how alkali metals and halogens are the most reactive elements because they are both only one electron away from a full outer shell.

After that we received handouts for a series of presentations on periodicity, and worked on that for a short while until the end of the session.