Currently unavailable: until 31/01/2016
Degree: D.Phil. in Organic Chemistry (Doctorate) - Oxford, Linacre College University
Hi I’m Peter.
I completed my undergraduate studies in Chemistry at the University of Warwick, where I was fortunate enough to graduate top of my year; having undertaken my Master’s research project in organic (carbon) chemistry. Following this, I won a place to continue my studies at doctoral level at the University of Oxford where I am now a final year Chemistry/Biology PHD student. Surprisingly I always found chemistry tricky at school, partly due to poor teaching, and had to teach myself (and others) the majority of AS/A2 chemistry. Since then I have continued informal and formal tutoring, including for GCSE Maths students, A-Level chemistry students and more recently as a 1st year tutor at the University of Oxford.
At school you’re often bombarded with often seemingly unconnected definitions and terms, which you’re then expected to apply to standardised situations. Unsurprisingly, this style of learning doesn’t appeal to many students. With any science subject it’s imperative to begin with understanding, often this will require me to draw in additional examples and explanations to illustrate the basic concepts which can then be used as a foundation for more advanced topics.
If students are keen and interested, I am also able to provide additional advanced background to the A2/AS course, topics that would be of interest for students interested in (and applying for) Chemistry, Medicine or Biology based subjects at University level.
The sessions should be guided by you and fun! Let me know what topics you struggle with and we can work through from the basics, worked examples and finally through past papers to ensure that the concepts are fully digested! By the end of the session, you should be teaching all your classmates!
If you have questions, feel free to mail me or book a “Meet the Tutor” session. Just let me know you’re exam board and what level you’re at!
|Biology||A Level||£26 /hr|
|Chemistry||A Level||£26 /hr|
|-Oxbridge Preparation-||Mentoring||£26 /hr|
|-Personal Statements-||Mentoring||£26 /hr|
|MChem in Chemistry||Masters Degree||1.1|
|A Level Mathematics||A-Level||A|
|A Level Chemistry||A-Level||A|
|A Level Biology||A-Level||A|
Krunal (Student) January 19 2015
Samantha (Student) January 17 2015
At a Basic Level
Entropy, whilst a mathematical concept can be described as a measure of disorder. The more disordered, then the greater the entropy. Correspondingly, a perfect crystal at 0K (absolute zero) has an entropy of 0, liquids will have considerably more and gases even more still. As an example, water in the solid state (ice) has a standard entropy of approximately 48 JK-1mol-1, water (liquid) approximately 70 JK-1mol-1 and water as steam, approximately 189 JK-1mol-1.
In Chemistry, Entropy is given the symbol S and the change in entropy ΔS; the units being JK-1mol-1. To calculate the entropy change for a reaction, one simply needs to utilise the following equation.
ΔS system = ∑S products - ∑S reactants (note the ∑ sign means the sum of)
In the calculation above, the entropy values are usually given as standard molar entropies (ΔS), that is the entropy at 298K (°C + 273 to convert to Kelvin).
Entropy can also be separated into 2 terms, the system (i.e. the reaction) and everything else or the universe.
ΔS total (or universe) = ΔS surroundings + ΔS system
Furthermore, the entropy change of the surroundings can also be calculated using:
ΔS surroundings = -ΔH reaction or system/T
For example, in an exothermic reaction (Enthalpy is negative). Energy, in the form of heat (kinetic energy) is given out to the surroundings. The result of this is to increase the disorder of the surroundings and thus the entropy of the surroundings increases. Even if the entropy of the system decreases, if the entropy of the surroundings increases to a greater extent, then the reaction will be feasible. This is a slightly long winded statement of the second law of thermodynamics
“for a reaction to occur then “ΔS total (or universe) > 0”
Entropy is at play in all chemical and physical processes: A gas has higher entropy than a solid, when a solid lattice dissolves entropy increases and when 1 mole of reagent makes 2 moles of products; entropy again increases.
Going back to the beginning with our definition of a perfect crystal having an entropic value of 0, we can now maybe think about this in a more detailed way (note this is advanced and for interest only). For a perfect crystal with no thermal energy (no movement), the atoms in the lattice are only arranged in one way or state. By using a famous equation (related to the Third Law), we can see how this arises. As W represents ways of arranging the particles (in this case W=1) and that ln(1)=0, we can see that S must therefore equal zero.
S = k*lnWsee more