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