Chem4 all notes

These contain all the notes from the book and I don't think I missed anything out other than the data you get given on the data sheet that I didn't need to write about

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  • Created on: 21-06-12 16:15
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CHEM4
Chapter 1.1 rate of chemical reactions
Rate of reaction is the change in concentration with unit time
To find the rate of reaction at an
instant, plot a graph with
concentration and time
and place a tangent at the curve and
find the gradient.
1.2 rate expression and order of reaction
Rate = K[A][B]
A and B both have the order 1 The overall order is 2
Rate = K[A][B]2
B has an order 2 which means it has double the effect on rate as A ­ The overall order is 3
The units of K depends on the overall order of a rate equation:
Rate = K[A][B] can be turned into
Rate = K which the units are Moldm3S1 The units on top
[A][B] Moldm3 x Moldm3 cancel out and the
bottom unit goes
above the divide and
( The unit S1 stays the same because it does not get so the units invert to
cancelled and is on top of the divide ) form Mol1dm3S1

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Determining the rate equation
To determine the order of each species in an equation plot a graph of concentration with
rate.
Small changes in temperature produces large changes
in reaction rates due to kinetic energy and activation energies.
1.4 Ratedetermining step
In a multistep reaction the rate determining step is the slowest step and one that contains all
the species in the rate equation with the ratios the same as the orders.
2.…read more

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Calculations using equilibrium constant expressions
Equation CH3OH CHOOH CHOOCH3 H2O
At start 1 mol 1 mol 0 mol 0 mol
At equilbrium (1x) mol (1x) mol x mol x mol
Kc = 4.0
Suppose at equilibrium the volume is Vdm3.…read more

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Doing the same to work out concentration of a reactant:
Equation CH3COCH3 HCN CH3C(CN)(OH)CH3
At start 4.00 mol X mol 0 mol
At equilbrium (4.00 1.00) (X 1) mol 1.00 mol
3.00 mol (X 1) mol
Volume of 2.00 dm3
Kc = 30.0 mol1dm3
CH3COCH3 = 3.00 / 2.00
HCN = (x ­ 1.00) / 2.00
CH3C(CN)(OH)CH3 = 1.00 / 2.00
Into the expression:
30.0 = 1.00 / 2.00 .
3.00 / 2.00 x (x ­ 1.00) / 2.00
30.0 x ( 3.00 / 2.…read more

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Reactants] concentration of products and reactants etc....
3.…read more

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H+]2
1.0 x 107 = [H+] or [OH]
3.2 The pH scale
pH = ­log10[H+]
­pH
from pH to H+ you have to do 10
Finding OH from pH
Say pH is 10
This means 1.0 x 1010 = [H+] we already know [H+] x [OH] = 1.0 x 1014
Therefore 1.0 x 1010 x [OH] = 1.0 x 1014
1.0 x 1014 = 1.0 x 104 which is the concentration of OH
1.…read more

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Finding pH of weak acids and bases
A weak acid equilibrium can be written as HA H+ + A
Put it into a constant: Kc = [H+] [A] but for an acid the symbol Kc is turned into Ka
[HA]
The larger the value of Ka the further the equilibrium is to the right and so the lower the pH is
(stronger the acid)
Example of calculating pH of weak acids
1.00 moldm3 of ethanoic acid Ka = 1.…read more

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Acidbase titrations
In a titration the equivalence point
indicated by the red line is where the
acid or alkali has just been
neutralised. However the point is not
always at pH 7.
You need to know CV/1000 = N
3.…read more

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The halfneutralisation point
Half way between zero and the equivalence point is the halfneutralisation point.
We know there is little change to the pH when we add acid (or a base) to the titration.
This helps us find the pKa of the weak acid
In the titre..
HA + OH H2o + A
At halfneutralisation point:
[HA] = [A] Therefore Ka = [H+] [A]
[HA]
Ka = [H+]
This means ­log10Ka = log10[H+]
So pKa = pH
3.…read more

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The function of having a salt component of a buffer is to act as a source of A ions to remove
any additional H+ ions.
Such as:
Na+A Na+ + A
The point of a buffer is to maintain the concentration of H+ ions and so pH. The [HA] and
[A] doesn't matter.…read more

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