Hi I'm Charlotte, a first year studying **Chemistry at Oxford University**.

I have always been **passionate about science** and when it came to applying for university, it was clear that Chemistry was the right choice for me. To my mind, everything comes down to Chemistry and it is this that makes the subject so fascinating.

**I enjoy teaching and have experience with tutoring** a Year 11 student through GCSE Maths and also my younger brother through his GCSE Science and Maths. I have previously given guidance to students in the year below me at school on AS level material.

At GCSE I studied a range of subjects and achieved 7A*'s and 3A's. I studied Chemistry, Biology and Maths at A level; getting 2A*'s, 1A and studied Physics at AS level (grade A). I got the highest in my year at school for pure A level Maths and full marks in my A2 Biology exams.

In the summer of 2012 I travelled to Venezuela for a month with World Challenge, where I got to exercise my **Spanish speaking skills** through booking travel and accommodation over the phone and communicating with the locals in everyday situations.

In my two terms at Oxford I have taken part in **Access events**, including a shadowing day where I took two Year 12 students to my lectures and showed them around the University. I have also just been elected as **Events Officer for the Chemistry Society**. This role involves approaching speakers and inviting them to come in and give talks on interesting topics.

Please don't hesitate to contact me with any questions.

Hi I'm Charlotte, a first year studying **Chemistry at Oxford University**.

I have always been **passionate about science** and when it came to applying for university, it was clear that Chemistry was the right choice for me. To my mind, everything comes down to Chemistry and it is this that makes the subject so fascinating.

**I enjoy teaching and have experience with tutoring** a Year 11 student through GCSE Maths and also my younger brother through his GCSE Science and Maths. I have previously given guidance to students in the year below me at school on AS level material.

At GCSE I studied a range of subjects and achieved 7A*'s and 3A's. I studied Chemistry, Biology and Maths at A level; getting 2A*'s, 1A and studied Physics at AS level (grade A). I got the highest in my year at school for pure A level Maths and full marks in my A2 Biology exams.

In the summer of 2012 I travelled to Venezuela for a month with World Challenge, where I got to exercise my **Spanish speaking skills** through booking travel and accommodation over the phone and communicating with the locals in everyday situations.

In my two terms at Oxford I have taken part in **Access events**, including a shadowing day where I took two Year 12 students to my lectures and showed them around the University. I have also just been elected as **Events Officer for the Chemistry Society**. This role involves approaching speakers and inviting them to come in and give talks on interesting topics.

Please don't hesitate to contact me with any questions.

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5from 2 customer reviews

Annabel (Student)

March 31 2015

Thank you so much Charlotte for all you help. I think I'm going to drop Chemistry, but I took the Cambridge Chemistry Challenge and got a bronze certificate, which was surprising! Thank you for all your help!

Annabel (Student)

April 18 2015

Thank you again for all your help!

Given a reaction in the form:

aA + bB ⇄ cC + dD

where A and B are reactants, C and D are products and a, b, c and d are stoichiometric coefficients (relative number of moles of each molecule).

The equilibrium coefficient is given by:

K_{c }= [C]^{c}[D]^{d} / [A]^{a}[B]^{b}

i.e. The concentration of each product raised to the power of its stoichiometric coefficient, divided by the concentration of each reactant raised to the power of its stoichiometric coefficient. **Note: it's the concentration of the products over reactants, not the reactants over products.**

e.g. H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

__Equilibrium constant units:__

The method for calculating units is similar to very simple algebra (though don't be put off if you never got on with this in school, it really is very simple). You can think of the units of concentration, mol dm^{-3}, as 'x'.

The first step is to replace the molecules in the square brackets with the units of concentration. Using our previous example:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (mol dm^{-3})^{2} / (mol dm^{-3})(mol dm^{-3})

You can see that the numerator, (mol dm^{-3})^{2}, is the same as the denominator, (mol dm^{-3})(mol dm^{-3}) and so all the units cancel. Therefore in this case, K_{c} has no units.

It may be easier to understand this if you think of mol dm^{-3} as 'x' and so instead of replacing the molecules in the square brackets with mol dm^{-3}, you replace them with x:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= x^{2}/ x x

= x^{2 }/ x^{2}

=^{ }no units

__Calculating the equilibrium constant:__

If you the information in the question gives you the equilibrium concentration then this step is very easy. **Note: the information must specify equilibrium concentration and not starting concentration.**

To do the calculation you simply plug in the equilibrium concentrations into your expression for Kc. For example for H_{2}(g) + I_{2}(g) ⇄ 2HI(g), equilibrium concentrations are:

H_{2 }= 0.125 mol dm^{-3}, I_{2 }= 0.020 mol dm^{-3}, HI = 0.500 mol dm^{-3}

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (0.500)^{2} / (0.125) x (0.020)

= 100 (no units)

If the question gives you starting concentrations and equilibrium concentrations of one component rather than equilibrium concentrations of all components, you can calculate equilibrium concentrations using an ICE table (initial amount/mol, change in amount/mol, equilibrium amount/mol. For each component of reaction). **Note: you must convert starting concentration to amounts (in moles) using container volume before putting them into the table. n = (concentration) x (volume)**

e.g. For: H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Starting amounts are: H_{2} = 0.35 mol, I_{2 }= 0.25 mol and equilibrium amount of HI = 0.40 mol. Volume of 1 dm^{3}.

Start by making an ICE table:

H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Initial/mol 0.35 0.25 0

Change/mol 0.40 (we know that we have gained 0.40mol HI)

Equilibrium/mol 0.40 (at this point we only know equilibrium amount of HI)

From here we know that HI has changed by +0.40 mol and also that there are two moles of HI in the overall equation. This means that H_{2} and I_{2 }amounts must change by half that. Also because they are our reactants, we must be losing amounts of them, so their changes in amount will be negative. So changes in H_{2} and I_{2} = -0.20 mol. To calculate equilibrium amount from here we simply add on this change to the initial amounts:

H_{2} = 0.35 + (-0.20) = 0.15 mol at equilibrium

I_{2} = 0.25 + (-0.20) = 0.05 mol at equilibrium

We can now plug these numbers into our table:

H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Initial/mol 0.35 0.25 0

Change/mol -0.20 -0.20 +0.40

Equilibrium/mol 0.15 0.05 0.40

Now we convert our equilibrium amounts into concentrations (using

n = (concentration) x (volume) --> (concentration) = n / (volume) ):

[H_{2}] = 0.15 mol / 1dm^{3 }= 0.15 mol dm^{-3}

[I_{2}] = 0.05 mol / 1dm^{3 }= 0.05 mol dm^{-3}

[HI] = 0.40 mol / 1dm^{3 }= 0.40 mol dm^{-3}

Finally all we need to do is put these equilibrium concentrations into our expression for equilibrium constant:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (0.40)^{2} / (0.15) x (0.05)

= 21.3 (no units)

Given a reaction in the form:

aA + bB ⇄ cC + dD

where A and B are reactants, C and D are products and a, b, c and d are stoichiometric coefficients (relative number of moles of each molecule).

The equilibrium coefficient is given by:

K_{c }= [C]^{c}[D]^{d} / [A]^{a}[B]^{b}

i.e. The concentration of each product raised to the power of its stoichiometric coefficient, divided by the concentration of each reactant raised to the power of its stoichiometric coefficient. **Note: it's the concentration of the products over reactants, not the reactants over products.**

e.g. H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

__Equilibrium constant units:__

The method for calculating units is similar to very simple algebra (though don't be put off if you never got on with this in school, it really is very simple). You can think of the units of concentration, mol dm^{-3}, as 'x'.

The first step is to replace the molecules in the square brackets with the units of concentration. Using our previous example:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (mol dm^{-3})^{2} / (mol dm^{-3})(mol dm^{-3})

You can see that the numerator, (mol dm^{-3})^{2}, is the same as the denominator, (mol dm^{-3})(mol dm^{-3}) and so all the units cancel. Therefore in this case, K_{c} has no units.

It may be easier to understand this if you think of mol dm^{-3} as 'x' and so instead of replacing the molecules in the square brackets with mol dm^{-3}, you replace them with x:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= x^{2}/ x x

= x^{2 }/ x^{2}

=^{ }no units

__Calculating the equilibrium constant:__

If you the information in the question gives you the equilibrium concentration then this step is very easy. **Note: the information must specify equilibrium concentration and not starting concentration.**

To do the calculation you simply plug in the equilibrium concentrations into your expression for Kc. For example for H_{2}(g) + I_{2}(g) ⇄ 2HI(g), equilibrium concentrations are:

H_{2 }= 0.125 mol dm^{-3}, I_{2 }= 0.020 mol dm^{-3}, HI = 0.500 mol dm^{-3}

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (0.500)^{2} / (0.125) x (0.020)

= 100 (no units)

If the question gives you starting concentrations and equilibrium concentrations of one component rather than equilibrium concentrations of all components, you can calculate equilibrium concentrations using an ICE table (initial amount/mol, change in amount/mol, equilibrium amount/mol. For each component of reaction). **Note: you must convert starting concentration to amounts (in moles) using container volume before putting them into the table. n = (concentration) x (volume)**

e.g. For: H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Starting amounts are: H_{2} = 0.35 mol, I_{2 }= 0.25 mol and equilibrium amount of HI = 0.40 mol. Volume of 1 dm^{3}.

Start by making an ICE table:

H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Initial/mol 0.35 0.25 0

Change/mol 0.40 (we know that we have gained 0.40mol HI)

Equilibrium/mol 0.40 (at this point we only know equilibrium amount of HI)

From here we know that HI has changed by +0.40 mol and also that there are two moles of HI in the overall equation. This means that H_{2} and I_{2 }amounts must change by half that. Also because they are our reactants, we must be losing amounts of them, so their changes in amount will be negative. So changes in H_{2} and I_{2} = -0.20 mol. To calculate equilibrium amount from here we simply add on this change to the initial amounts:

H_{2} = 0.35 + (-0.20) = 0.15 mol at equilibrium

I_{2} = 0.25 + (-0.20) = 0.05 mol at equilibrium

We can now plug these numbers into our table:

H_{2}(g) + I_{2}(g) ⇄ 2HI(g)

Initial/mol 0.35 0.25 0

Change/mol -0.20 -0.20 +0.40

Equilibrium/mol 0.15 0.05 0.40

Now we convert our equilibrium amounts into concentrations (using

n = (concentration) x (volume) --> (concentration) = n / (volume) ):

[H_{2}] = 0.15 mol / 1dm^{3 }= 0.15 mol dm^{-3}

[I_{2}] = 0.05 mol / 1dm^{3 }= 0.05 mol dm^{-3}

[HI] = 0.40 mol / 1dm^{3 }= 0.40 mol dm^{-3}

Finally all we need to do is put these equilibrium concentrations into our expression for equilibrium constant:

K_{c }= [HI]^{2} / [H_{2}] [I_{2}]

= (0.40)^{2} / (0.15) x (0.05)

= 21.3 (no units)