FINAL FINAL FINAL DECK

Question 1

What does 3 bonded pairs around the central atom produce?

Answer 1

Trigonal planar

120 degrees

Question 2

3 bonded pairs and one lone pair around central atom?

Answer 2

Trigonal pyramidal

Lone pair at the top, everything bellow same structure as tetrahedral

107 degree bond angle

Question 3

2 bonded pairs and 2 lone pairs around central atom?

Answer 3

nonlinear shape (V shaped) 
104.5 degrees

Question 4

5 bonded pairs around central atom?

Answer 4

Trigonal bipryamidal
Around central atom there is a straight line going through, and perpindicular to this there is a triangle of atoms (use both wedges in this)
Producing 90 degree angles and 120 degree angles)

Question 5

Formula for moles in gas?

Answer 5

Moles = volume / 24dm^3

Question 6

However there are 2 exceptions in the rule that ionisation energy increases along a period between group 2 to 3 and group 5 to 6 explain why?

Answer 6

The outer electrons in group 3 elements is a p orbital, rather than an S orbital, which means it’s further away from the nucleus, and has additional shielding from s orbital, theses factors overide the increased nuclear charge

In the group 5 elements, the electron is being removed from a singly-occupied orbital, in group 6 being removed from orbital containing 2 electrons. The repulsion means between 2 electrons in a orbital makes it easy to remove from shared orbitals

Question 7

What’s a chlorate ion?

Answer 7

ClO-

Question 8

How do you make bleach?

Answer 8

React chlorine gas with cold dilute aqueous sodium hydroxide in a disproportionation reaction

Question 9

How is chlorine used to kill bacteria in water?

Answer 9

Cl2 + H2O =(reversible reaction) HCl + HClO

HClO + H2O = (reversible reaction) ClO- + H3O+

Chlorate ions kill bacteria

Question 10

What order should you do tests to avoid false negatives?

Answer 10

Test for carbonates - no CO2, so no barium carbonate precipitate can form
Test for sulphates - no sulphate to react with silver nitrate to form silver sulphate
Test for halides

Question 11

What happens to KBr (colourless) when in aqueous or organic solution and Chlorine water, Bromine water, and Iodine solution are added?

Answer 11

In aqueous solution:
Chlorine water added: Goes yellow from Br2
Bromine water added: no reaction
Iodine solution added: no reaction

In organic solution:
Chlorine water added: Goes orange from Br2
Bromine water added: no reaction
Iodine solution added: no reaction

Question 12

What happens to KI (colourless) when in aqueous or organic solution and Chlorine water, Bromine water, and Iodine solution are added?

Answer 12

In aqueous solution:
Chlorine water added: Goes orange/brown as I2 displaced
Bromine water added: Goes orange/brown as I2 displaced
Iodine solution added: no reaction

In organic solution:
Chlorine water added: Goes purple as I2 displaced
Bromine water added: Goes purple as I2 displaced
Iodine solution added: no reaction

Question 13

What is enthalpy change in terms of bond forming and breaking?

Answer 13

The overall affect of both of them occuring

Average bond enthalpy of breaking bonds - average bond enthalpy of forming bonds

Question 14

How can enthalpy changes be worked out using the enthalpy change of combustion?

Answer 14

Same method before, however the direction of arrows have changed and so has what’s on the bottom

On the bottom have water and CO2 and both the lines will be going downwards with an amount of oxygen next to them to balance the equation

Work out total distance taking into account second line is going to wrong way

Question 15

What is the reaction rate?

Answer 15

The change in the amount of reactants or products per unit time
Units are mol dm^-3 s^-1

Question 16

Explain the iodine clock reaction?

Answer 16

H2O2 (aq) + 2I- (aq) + 2H+ (aq) = 2H2O (l) + I2(g)

Small amount of Sodium Thiosulfate solution and starch are added to an excess of H2O2 (Hydrogen peroxide), and Iodide ions in acid solution (H+)

Starch is an indicator and turns blue/black in presence of iodine

Sodium thiosulfate instantly reacts with any Iodine that forms making it go back to iodide ions

So once the set amount of Sodium thiosulfate is used up solution will go blue/black, so can vary iodide or H2O2 concs to give you different times

Question 17

All the graphs for conc-time and rate conc?

Answer 17

Conc-time
0- diagonal straight line from y axis
1st Curved line with diagonal gradient from y axis, reaching an asymptote near x axis

Rate conc -
0-Horizontal line from Y axis
1st - Positive gradient for y=x, through the origin

2nd - Positive curve going through the origin, Rate is proportional to concentration^2

Question 18

Half life for 1st order reactant and second order?

Answer 18

1st - constant

2nd not constant

Question 19

How do you work out the rate constant of a first order reaction if you know it’s half life?

Answer 19

k = ln2 / half life

Units = s^-1

Question 20

How do you find the total pressure of a gas mixture?

Answer 20

It’s the sum of all the partial presures

Question 21

How do you find the partial pressure of a gas?

Answer 21

First find mole fraction of the gas in the mixture using Number of moles of gas / total number of all moles of gas in mixture

The multiply the mole fraction of gas by the total pressure of the mixture

So if they have the same amount of moles in equation will have the same partial pressure

Question 22

What’s the equation for Kp?

Answer 22

if aA(g) + bB(g) ⇋ cC(g) + dD(g)

Then Kp = p(C^c) x p(D^d) / p(A^a) x p(B^b)

Use the partial pressures instead of concentrations

Units are kPa

NOT SQUARE BRACKETS

Question 23

Formula to find pKa?

Answer 23

pKa = -log10 Ka

remember that it’s weird one and you are finding it out the same way for pH

Question 24

Formula to find Ka when you know pKa?

Answer 24

Ka = 10^-pKa

Question 25

Describe features f Methyl orange indicator?

Answer 25

Colour at low pH is red Colour at high pH is yellow Approx pH of colour change is between 3 and 4.5 So useful for strong acid/ strong base and strong acid/ weak base

Question 26

Describe phenolphthallein?

Answer 26

Colour at low pH is colourless Colour at high pH is pink Approx pH of colour change is 8 to 10 So useful for strong acid/ strong base and weak acid/ strong base

Question 27

How do you calculate entropy?

Answer 27

Entropy = Entropy of products - Entropy of reactants So entropy has to increase during a reaction for it to be feesible

Question 28

What are the 2 half equations for the redox titration of Acidified Potassium manganate oxidising Fe(2+) to Fe(3+) and the colour change and what does this titration tell us?

Answer 28

MnO4(-) + 8H(+) + 5e- = Mn(2+) + 4H2O 5Fe(2+) = 5Fe(3+) + 5 e- So equals MnO4(-) + 8H(+) + 5Fe(2+) = Mn(2+) +4H2O + 5Fe(3+) Once the reaction has finished will see a colour change from purple to colourless as MnO4(-) is pink and {Mn(H2O)6}(2+) is colourless How much Oxidising agent we need to reduce the species ``` Can work out both half equations by: MnO4 - = Mn(2+) MnO4- = Mn(2+) + 4H2O MnO4- + 8H+ = Mn(2+) + 4 H2O MnO4- + 8H+ + 5e- = Mn(2+) +4H2O ``` ``` Fe(2+) = Fe(3+) Fe(2+) = Fe(3+) + e- 5Fe(2+) = 5Fe(3+) + 5e- ```

Question 29

What are the 2 half equations for Potassium dichromate solution K2Cr2O7 oxidising Zn to Zn(2+) and the colour change and what does this titration tell us?

Answer 29

Cr2O7(2-) + 14H+ + 6e- = 2Cr(3+) + 7H2O 3Zn = 3Zn(2+) +6e- So equals Cr2O7(2-) + 14H+ + 3Zn = 2Cr(3+) + 7H2O + 3Zn(2+) Colour change when reaction is finished will be orange to green as Cr2O7(2-) is orange and {Cr(H2O)6}(3+) looks green How much Oxidising agent we need to reduce the species ``` Can work out by: Cr2O7(2-) = 2Cr(3+) Cr2O7(2-) = 2Cr(3+) + 7H2O Cr2O7(2-) +14H+ = 2Cr(3+) + 7H2O Cr2O7(2-) +14H+ +6e- = 2Cr(3+) + 7H2O ``` ``` Zn = Zn(2+) Zn = Zn(2+) +2e- 3Zn = 3Zn(2+) + 6e- ```

Question 30

All the steps and equations for a sodium thiosulfate titration to find the concentration of an oxidising agent?

Answer 30

Use a certain volume Potassium Iodate solution to oxidise an excess of KI IO3(-) + 5I(-) + 6H+ = 3I2 + 3H2O To work out: IO3(-) 5I(-) = 3I2 IO3(-) 5I(-) = 3I2+3H2O IO3(-) 5I(-) +6H(+) = 3I2+3H2O Find out how many moles of iodine have been produced, by titrating the resulting solution with a known concentration of Sodium Thiosulfate (Na2S2O3), when the solution goes pale yellow showing it's near to the end point, add starch solution, which will show that Iodine is still present by being blue/black, will know reaction is finished when blue/black dissapears I2 + 2S2O3(2-) = 2I- + S4O6(2-) Find the moles of Iodine from second equation, then input it into first equation to find moles of oxidising agent ( divide by 3) and then can find concentration