I am a third year Chemistry undergraduate at Durham University looking to tutor Chemistry A-level students or Maths and Chemistry at GCSE. I achieved A* grades in both Chemistry and Mathematics at A-level and want to help others attain the best they possibly can.

I have tutored for two years with my tutor as well as during A level and so know many of the common pitfalls in understanding which people have. I am told by tutees that my explanations are always in depth and easy to understand.

**What experience do I have to be a good tutor?**

I have been tutoring Chemistry for two years both with private tutees and through the schools programme and so have experience of explaining and teaching commonly challenging concepts such as chromatography and reaction kinetics, in a way you'll understand.

Through tutoring A-level and Cambridge Pre-U qualifications I am experienced at explaining and teaching high level concepts and am always prepared to explain concepts in further detail if I feel it will aid understanding.

I am a third year Chemistry undergraduate at Durham University looking to tutor Chemistry A-level students or Maths and Chemistry at GCSE. I achieved A* grades in both Chemistry and Mathematics at A-level and want to help others attain the best they possibly can.

I have tutored for two years with my tutor as well as during A level and so know many of the common pitfalls in understanding which people have. I am told by tutees that my explanations are always in depth and easy to understand.

**What experience do I have to be a good tutor?**

I have been tutoring Chemistry for two years both with private tutees and through the schools programme and so have experience of explaining and teaching commonly challenging concepts such as chromatography and reaction kinetics, in a way you'll understand.

Through tutoring A-level and Cambridge Pre-U qualifications I am experienced at explaining and teaching high level concepts and am always prepared to explain concepts in further detail if I feel it will aid understanding.

**From the basics**

I have always found that the best way of understanding something is from the basics. This especially applies in science whereby you will constantly have to build on and apply concepts that you have learned. I will strive in my tutorials to ensure that you understand the basics of a concept as with these solid foundations it becomes much easier to grasp more complicated ideas and apply them to exam questions. Having studied degree level chemistry for three years I have deep understanding of the subject and can explain concepts to you in ways which have worked for me in the past.

**Tailored around you**

I know the areas which I struggled with understanding whilst studying for my exams. My tutorials however will be tailored around you.

Of course, I will explain the common pitfalls I encountered in grasping a concept but precisely which bits of the course we cover depends on what you feel you need. This may be a small area of the course or the whole syllabus. I can then plan around this to ensure you get the most value out of our time together.

**Exam boards**

I studied OCR Chemistry B (Salters) at A level however if you study under another exam board please contact me, I’m sure the content will be very similar.

For GCSE much of the content is universal. Please contact me with your exam board nonetheless so that I can look over the syllabus beforehand.

**From the basics**

I have always found that the best way of understanding something is from the basics. This especially applies in science whereby you will constantly have to build on and apply concepts that you have learned. I will strive in my tutorials to ensure that you understand the basics of a concept as with these solid foundations it becomes much easier to grasp more complicated ideas and apply them to exam questions. Having studied degree level chemistry for three years I have deep understanding of the subject and can explain concepts to you in ways which have worked for me in the past.

**Tailored around you**

I know the areas which I struggled with understanding whilst studying for my exams. My tutorials however will be tailored around you.

Of course, I will explain the common pitfalls I encountered in grasping a concept but precisely which bits of the course we cover depends on what you feel you need. This may be a small area of the course or the whole syllabus. I can then plan around this to ensure you get the most value out of our time together.

**Exam boards**

I studied OCR Chemistry B (Salters) at A level however if you study under another exam board please contact me, I’m sure the content will be very similar.

For GCSE much of the content is universal. Please contact me with your exam board nonetheless so that I can look over the syllabus beforehand.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

Enhanced DBS Check

15 Sep, 2016In chemical reactions entropy can be considered to be a measure of the number of ways which molecules and their quanta of energy can be arranged and is measured in J K^{-1} mol^{-1}.

For a process to occur spontaneously the total entropy must always increase (This is the second law of thermodynamics). The entropy change of a reaction can be used to decide whether or not a reaction will occur spontaneously at a certain temperature. Bear in mind that this gives no indication of how quickly a reaction can occur only that it will; the reaction rate may be so low that the reaction isn’t noticeable/measurable.

In general **solids have lower entropies than liquids which have lower entropies than gases**. This can be explained very simply by looking at the molecular arrangements in each phase:

- In a solid the molecules are usually arranged in some form of lattice structure whereby movement of the molecules is severely restricted. This means that there are very few ways which the molecules can be arranged in the space and so the entropy is low.

- In a liquid the molecules are arranged randomly and are less restricted in their movements, this therefore means that they have higher entropies than solids as the molecules can be arranged more ways in the space.

- In gases the molecules are also randomly distributed in the space and are even more diffuse than in liquids with very little restriction to their movements in comparison. There are therefore more ways than in either solids or liquids for the molecules to be arranged in the space and so there entropies are generally the highest.

**Example**

You can use the above rules to make a very basic/simplified decision on whether a reaction will occur spontaneously just by looking at the relative phases of the reactants and products, for example in the reaction below:

A_{(l) }+ B_{(l)} → 2C_{(g)}

Two liquids react to give a gas, gases have higher entropies than liquids and so by the above rules the entropy change of this reaction will be positive and so it will occur spontaneously.

**Considering the surroundings**

So far we have considered the entropy change of the system (**ΔS _{system} = ΣS_{products} - ΣS_{reactants}**). What hasn’t been considered here however is that entropy is also a measure of the arrangement of quanta of energy not just the distribution of molecules. The energy changes associated with a process must also therefore be factored in.

An equation you may have come across is **ΔS _{surroundings} = -ΔH/T**. This equation allows us to consider the entropy change of the surroundings during a process rather than just that of the system. It is calculated by dividing the negative of the enthalpy change of the process by the temperature.

- This formula basically allows us to consider the distribution of the quanta of energy rather than just that of the molecules. It thus explains why processes occur which would seem impossible by just considering the differences between the starting and final molecular distributions. An example of this may be the freezing of water at -10^{o}C. The idea of a liquid spontaneously turning into a solid would seem impossible by just considering the molecular distribution. It is moving from a less ordered to a more ordered state, however by considering the total entropy change of the process it can be rationalised.

The total entropy change of a reaction is calculated by **ΔS _{total} = ΔS_{system} - ΔS_{surroundings}**. Using this equation you can ascertain whether a reaction will occur spontaneously at a given temperature.

**Example**

An example of this is the freezing of water. This is a process, the temperature dependence of which we come across on a daily basis (water freezes at 0^{o}C or 273K). This temperature dependence can be seen very clearly by looking at the total entropy change of the process **(ΔS _{total} = ΔS_{system} + ΔS_{surroundings})**. The enthalpy change of water freezing is about

This means that **ΔS _{surroundings} = -ΔH/T = - (-6010 J mol^{-1 }/ T) = +6010 J mol^{-1 }/ T**

For water freezing **ΔS _{system} =**

Therefore **ΔS _{total} = ΔS_{system} + ΔS_{surroundings} = -22.0 J K^{-1} mol^{-1} + (+6010 J mol^{-1 }/ T)**

**= -22.0 J K ^{-1} mol^{-1} + 6010 J mol^{-1 }/ T).**

At 263K (-10^{o}C) **ΔS _{total} = -22.0 J K^{-1} mol^{-1} + (6010 J mol^{-1 }/ 263K) **

**= -22.0 J K ^{-1} mol^{-1} + (6010 J mol^{-1 }/ 283K) = -22.0 J K^{-1} mol^{-1} + 22.9 J K^{-1} mol^{-1 }= +0.9 J mol^{-1 }K^{-1}**

The entropy change of the process is positive and so the process will occur spontaneously at this temperature (water does freeze at -10^{o}C).

If we look at whether water will freeze at 283K (+10^{o}C) however, we get a different story:

**ΔS _{total} = ΔS_{system} + ΔS_{surroundings} = -22.0 J K^{-1} mol^{-1} + (+6010 J mol^{-1 }/ T)**

**ΔS _{total} = -22.0 J K^{-1} mol^{-1} + (6010 J mol^{-1 }/ 283K)**

**= -22.0 J K ^{-1} mol^{-1} + (6010 J mol^{-1 }/ 283K) = -22.0 J K^{-1} mol^{-1} + 21.2 J K^{-1} mol^{-1 }= -0.8 J K^{-1} mol^{-1}**

The entropy change of the process of water freezing at 283K is negative, therefore this process will not occur spontaneously (water doesn’t freeze at +10^{o}C).

As you can see, using entropy calculations we can rationalise the temperature at which processes happen, the freezing of water in this case. These calculations can also be used to find the minimum/threshold temperature at which a process will occur spontaneously (the temperature whereby **ΔS _{total} = 0**). This is the temperature at which the phase change takes place, the point of equilibrium between the phases.

**At higher level…**

If you study chemistry to a higher level you will find that these calculations are simplified. There are also entropy changes associated with the difference in the temperature that the process takes place (-/+10^{o}C) and that at which the phase change takes place (0^{o}C). The diagrams used for mapping out the total change in entropy for the process, taking these extra changes into account, are similar to the Hess cycles you may have already come across.

All data for the calculations of the entropy changes of water freezing used from *Chemical Ideas Third Edition (2008) Pearson Education Limited 2008 ISBN 978 0 435631 49 9*

In chemical reactions entropy can be considered to be a measure of the number of ways which molecules and their quanta of energy can be arranged and is measured in J K^{-1} mol^{-1}.

For a process to occur spontaneously the total entropy must always increase (This is the second law of thermodynamics). The entropy change of a reaction can be used to decide whether or not a reaction will occur spontaneously at a certain temperature. Bear in mind that this gives no indication of how quickly a reaction can occur only that it will; the reaction rate may be so low that the reaction isn’t noticeable/measurable.

In general **solids have lower entropies than liquids which have lower entropies than gases**. This can be explained very simply by looking at the molecular arrangements in each phase:

- In a solid the molecules are usually arranged in some form of lattice structure whereby movement of the molecules is severely restricted. This means that there are very few ways which the molecules can be arranged in the space and so the entropy is low.

- In a liquid the molecules are arranged randomly and are less restricted in their movements, this therefore means that they have higher entropies than solids as the molecules can be arranged more ways in the space.

- In gases the molecules are also randomly distributed in the space and are even more diffuse than in liquids with very little restriction to their movements in comparison. There are therefore more ways than in either solids or liquids for the molecules to be arranged in the space and so there entropies are generally the highest.

**Example**

You can use the above rules to make a very basic/simplified decision on whether a reaction will occur spontaneously just by looking at the relative phases of the reactants and products, for example in the reaction below:

A_{(l) }+ B_{(l)} → 2C_{(g)}

Two liquids react to give a gas, gases have higher entropies than liquids and so by the above rules the entropy change of this reaction will be positive and so it will occur spontaneously.

**Considering the surroundings**

So far we have considered the entropy change of the system (**ΔS _{system} = ΣS_{products} - ΣS_{reactants}**). What hasn’t been considered here however is that entropy is also a measure of the arrangement of quanta of energy not just the distribution of molecules. The energy changes associated with a process must also therefore be factored in.

An equation you may have come across is **ΔS _{surroundings} = -ΔH/T**. This equation allows us to consider the entropy change of the surroundings during a process rather than just that of the system. It is calculated by dividing the negative of the enthalpy change of the process by the temperature.

- This formula basically allows us to consider the distribution of the quanta of energy rather than just that of the molecules. It thus explains why processes occur which would seem impossible by just considering the differences between the starting and final molecular distributions. An example of this may be the freezing of water at -10^{o}C. The idea of a liquid spontaneously turning into a solid would seem impossible by just considering the molecular distribution. It is moving from a less ordered to a more ordered state, however by considering the total entropy change of the process it can be rationalised.

The total entropy change of a reaction is calculated by **ΔS _{total} = ΔS_{system} - ΔS_{surroundings}**. Using this equation you can ascertain whether a reaction will occur spontaneously at a given temperature.

**Example**

An example of this is the freezing of water. This is a process, the temperature dependence of which we come across on a daily basis (water freezes at 0^{o}C or 273K). This temperature dependence can be seen very clearly by looking at the total entropy change of the process **(ΔS _{total} = ΔS_{system} + ΔS_{surroundings})**. The enthalpy change of water freezing is about

This means that **ΔS _{surroundings} = -ΔH/T = - (-6010 J mol^{-1 }/ T) = +6010 J mol^{-1 }/ T**

For water freezing **ΔS _{system} =**

Therefore **ΔS _{total} = ΔS_{system} + ΔS_{surroundings} = -22.0 J K^{-1} mol^{-1} + (+6010 J mol^{-1 }/ T)**

**= -22.0 J K ^{-1} mol^{-1} + 6010 J mol^{-1 }/ T).**

At 263K (-10^{o}C) **ΔS _{total} = -22.0 J K^{-1} mol^{-1} + (6010 J mol^{-1 }/ 263K) **

**= -22.0 J K ^{-1} mol^{-1} + (6010 J mol^{-1 }/ 283K) = -22.0 J K^{-1} mol^{-1} + 22.9 J K^{-1} mol^{-1 }= +0.9 J mol^{-1 }K^{-1}**

The entropy change of the process is positive and so the process will occur spontaneously at this temperature (water does freeze at -10^{o}C).

If we look at whether water will freeze at 283K (+10^{o}C) however, we get a different story:

_{total} = ΔS_{system} + ΔS_{surroundings} = -22.0 J K^{-1} mol^{-1} + (+6010 J mol^{-1 }/ T)

**ΔS _{total} = -22.0 J K^{-1} mol^{-1} + (6010 J mol^{-1 }/ 283K)**

**= -22.0 J K ^{-1} mol^{-1} + (6010 J mol^{-1 }/ 283K) = -22.0 J K^{-1} mol^{-1} + 21.2 J K^{-1} mol^{-1 }= -0.8 J K^{-1} mol^{-1}**

The entropy change of the process of water freezing at 283K is negative, therefore this process will not occur spontaneously (water doesn’t freeze at +10^{o}C).

As you can see, using entropy calculations we can rationalise the temperature at which processes happen, the freezing of water in this case. These calculations can also be used to find the minimum/threshold temperature at which a process will occur spontaneously (the temperature whereby **ΔS _{total} = 0**). This is the temperature at which the phase change takes place, the point of equilibrium between the phases.

**At higher level…**

If you study chemistry to a higher level you will find that these calculations are simplified. There are also entropy changes associated with the difference in the temperature that the process takes place (-/+10^{o}C) and that at which the phase change takes place (0^{o}C). The diagrams used for mapping out the total change in entropy for the process, taking these extra changes into account, are similar to the Hess cycles you may have already come across.

All data for the calculations of the entropy changes of water freezing used from *Chemical Ideas Third Edition (2008) Pearson Education Limited 2008 ISBN 978 0 435631 49 9*