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BENCHMARK PRACTICE EXAM ONE
ANSWERS BONDING AND THERMOCHEMISTRY ANSWERS
QUESTION ONE
(a) Complete the following table by
Molecule Lewis diagram Name of shape Polar or non-polar
Cl 2 O Bent
Polar
BCl 3 Triangular planar Non-polar
NCl 3
Trigonal pyramid Polar
CCl 4 tetrahedral Non-polar
Explain why each of the molecules BCl 3 and Cl 2 O have the shape you identified in part (a) above.
(i) Cl 2 O
There are 4 areas of electron density around the central atom O. These areas occupy positions to minimize repulsion OR repel maximally to form a tetrahedral arrangement. As 2 areas are bonding and 2 are non- bonding, the shape observed is BENT.
3 lines correct = m point
3 shapes correct = a point
3 Lewis diagrams correct = a point
3 polarities correct = a point
(ii) BCl 3
There are 3 areas of electron density around the central atom B. These areas occupy positions to minimize repulsion OR repel maximally to form a triangular planar arrangement. As all 3 areas are bonding, the shape observed is TRIANGULAR PLANAR.
(c) Justify your classification of polar or non-polar in part (a) for each of the molecules BCl 3 and Cl 2 O
(i) Cl 2 O
There are polar bonds in this molecule as O is more electronegative than Cl. Due to the bent shape, these dipoles are arranged asymmetrically around the central atom, and so the dipoles do not cancel and this molecule is POLAR.
(ii) BCl 3
There are polar bonds in this molecule as Cl is more electronegative than B. Due to the triangular planar shape, these dipoles are arranged symmetrically around the central atom, and so the dipoles do cancel and this molecule is NON-POLAR.
N0 N1 N2 A3 A4 M5 M6 E7 E
Nothing correct
1 a point 2 a points 3a points 4 a points 2 m points
3 m points
1 e point (minor error)
1e point
One statement correct for shape (correct # areas bonding and not, repulsion etc) = 1 a point One statement correct for polarity (eg polar bonds described with reason or cancelling dipoles due to symmetry etc) = 1 a point One shape explained correctly = 1 m point One polarity explained correctly = 1 m point Both shapes and polarities correct = 1 e point
QUESTION THREE
Iron metal and the compound iron(II) chloride are distinctly different substances. Some physical properties of each are given below.
Iron metal Iron(II) chloride High melting point High melting point
An excellent conductor of electricity in both the solid and molten state.
The solid does not conduct electricity but does in the molten state.
Discuss the similarities and differences in the given properties in these two substances, in terms of the types of particles present and the structure and bonding within the substance.
Fe is a metallic substance. It is a 3-d lattice composed of Fe atoms bonded to each other by strong metallic bonds. The valence e are delocalized /the atoms are in a sea of delocalized valence electrons. As the metallic bonds are strong, it requires a lot of energy to overcome these bonds and convert solid Fe to a liquid, therefore Fe has a high melting point. In the solid structure there are free moving valence electrons (charged), therefore solid and liquid Fe can conduct electricity.
FeCl 2 is an ionic substance. It is a 3-d lattice composed of Fe2+^ and Cl-^ ions bonded to each other by strong ionic bonds (electrostatic attraction between positive and negative charges). As the ionic bonds are strong, it requires a lot of energy to overcome these bonds and convert solid FeCl 2 to a liquid, therefore FeCl 2 has a high melting point. In the solid structure the charged ions are locked into the 3D lattice and are unable to move so solid FeCl 2 cannot conduct electricity. However, when molten, the ions are free to move and liquid FeCl 2 can conduct electricity.
N0 N1 N2 A3 A4 M5 M6 E7 E
Nothing correct
1 a point 2 a points 3a points 4 a points 1 m points
2 m points
1 e point 2e point
conductivity correctly defined = a point
partial explanation of one property (not enough for merit) = a point
One substance melting point explained correctly = m point
One substance conductivity correctly explained = m point
One substance explained correctly for BOTH conductivity and melting point = e point
QUESTION FOUR
Amino acids are the building blocks of protein. The simplest amino acid is glycine (aminoethanoic acid). The structure of glycine is drawn below, with no attempt to show bond angles correctly.
(a) (i) Complete this diagram by adding the non-bonding electron pairs.
2 e added to the N atom, 2 PAIRS (4e) added to the O atom on the end (to make sure all non- bonding e are used)
Bond Bond angle
(iii) Explain carefully how you made your choices for the bond angles above.
Bond angles depend only on the number of areas of e density around the central atom, whether bonded or non-bonded. As these areas repel maximally, the angles are 109.5 for 4 areas of e density and 120 for 3 areas of e density.
(b) The structure of a compound can affect it’s solubility in water. Using CO 2 and HBr as examples explain why one of these is soluble in water and one is not.
e added correctly = 1a point
(c) Calculate the energy change for the above reaction when 3 mol of HI is produced
2mol HI absorbs 52 kJ (from equation), therefore
1 mol HI absorbs 28 kJ and
3 mol HI absorbs 3 x 28 = 84 kJ
(d) What mass of HI would be produced when 364 kJ of energy is absorbed in the above reaction? M (HI) = 128 gmol-
1 mol HI absorbs 52 kJ therefore
X mol HI absorbs 364 kJ
X = 364/52 = 7 mol. M = nM = 7 x 128 = 896 g of HI
BENCHMARK PRACTICE ANSWERS REACTIONS ANSWERS
QUESTION ONE
The following reaction was investigated experimentally. The total volume of each experiment
was 200 mL. The investigation and results are shown in the table below:
S 2 O 8 2-^ + 2I-^ → 2SO 4 2-^ + I 2
Time (minutes)
Mass of I 2 produced (g)
0.1 mol S 2 O 82 -^ 0.2 mol S 2 O 82 -^ 0.4 mol S 2 O 82 -^ 0.1 mol S 2 O 82 -^ + Fe2+
2 0.02 0.02 0.03 0.
4 0.03 0.04 0.06 0.
6 0.04 0.06 0.06 0.
8 0.05 0.06 0.06 0.
(a) What factor affecting reaction rate is being investigated in the first 3 columns? Concentration
(b) What is the most likely factor affecting reaction rate being investigated in the 4th^ column? Catalyst
(c) Write one conclusion that can be made about reaction rate from the table.
The higher the concentration, the faster the rate of reaction OR use of a catalyst increases the rate of reaction
(d) Explain carefully how BOTH factors being investigated affect reaction rate with reference to the collision of particles.
An increased concentration means more particles in the same volume. Therefore there is a higher chance of a successful collision as there are more collisions and so a higher frequency of effective collisions.
A catalyst reduces the Ea by providing a lower energy alternative path. This means that more particles will have sufficient energy to overcome the Ea barrier and therefore ther will be more successful collisions per second
QUESTION TWO
Ammonia is made by the Haber Process.
N 2 (g) + 3H 2 (g) 2NH 3 (g) ∆ H = – 92 kJ mol–^1
(a) Write down the form of the equilibrium constant expression for this equilibrium.
(b) An industrial plant to produce ammonia is run at 400 °C. Discuss how both rate and equilibrium factors of concentration, pressure and temperature need to be considered when deciding the ‘best’ temperature for this process.
Equilibrium: To increase the conc of NH 3 , you could decrease the temperature, increase the pressure and decrease the conc of NH 3 (remove it as it forms) and/or increase the conc of either of the 2 reactants. By decreasing the temp, the system moves to the side to increase the temp – the exothermic direction. As DH is negative, this is the forward direction, increasing the conc of NH3. Increasing the pressure produces a shift to the side with the fewest moles of gas particles – in this case the right (2 moles compared to the left has 4 moles). This increases the conc of NH3. Increasing the conc of reactants or decreasing the conc of products both produce a shift to the right to make more NH3.
Rate: decreasing the temperature decrease the Ek of the particles, resulting in fewer particles having sufficient energy to overcome Ea barrier and therefore a slower rate of reaction (decreasing the production of NH3 in the same time frame). Increasing the conc of reactants would increase the rate, more particles per mL, means a higher frequency of effective collisions.
Therefore, you would need a low enough temp to move the equilibrium to the right without decreasing the rate too much, a high conc of reactants and a high pressure
(c) The following reaction is exothermic: 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g)
Both N 2 O 5 and O 2 are colourless gases and NO 2 is a brown gas. A mixture of these gases exists at equilibrium and is observed as a brown colour. (i) Complete the equilibrium constant expression for the reaction.
(c) Place the following solutions in order of increasing pH, by writing them into the boxes below.
0.01 mol L–^1 CH 3 COOH 0.01 mol L–^1 HCl 0.1 mol L–^1 HCl 0.1 mol L–^1 NaOH
0.1 HCl 0.01 HCl 0.01 CH 3 COOH 0.1 NaOH
Lowest pH Highest pH
Justify your order above in terms of: proton transfer relative concentrations of [H 3 O+] and [OH–], linked to the pH of the solution.
You should include equations in your answer.
HCl + H 2 O ↔ H 3 O+ + Cl-
CH 3 COOH + H 2 O ↔ CH 3 COO-^ + H 3 O+
NaOH → Na+ + OH-
HCl is a strong acid which fully dissociates into its ions (see equation 1). The 0.1 molL-1^ HCl is
also the most concentrated, so this sample has the highest [H 3 O+] and therefore (as pH = - log
[H 3 O+] this sample has the lowest pH. 0.01 molL-1 is also fully dissociated, but as this is less
concentrated, the [H 3 O+] is slightly lower and therefore the pH is a little higher.
CH 3 COOH is a weak acid which partially dissociates into its ions (see equation 2). 0.01 molL-
is the same concentration as the 2nd^ HCl, but as this acid is weak, the [H 3 O+] is slightly lower
and therefore the pH is a little higher, therefore the 3rd^ highest pH.
All of these acids produce H 3 O+^ so they are all acidic (pH<7)
NaOH is a strong base and fully dissociates into its ions (see equation 3). Since it produces a
high concentration of OH- (which is made in the reaction) as it is both strong and concentrated,
and therefore the lowest [H 3 O+] (as [H 3 O+] x [OH-] = 1 x 10-14) this has the highest pH
BENCHMARK PRACTICE ANSWERS ORGANIC ANSWERS
QUESTION ONE
(a) Complete the following table.
Structure Name
CH 3 CH 2 CH(CH 3 )CH=CH 2 (i) 3-methylpent-1-ene
(ii) CH 3 CHClC(CH 3 ) 2 CH 3 2-chloro-3,3-dimethylbutane
CH 3 CH(Cl)CH 2 CH 2 COOH (iii) 4-chloropentanoic acid
(iv) C(OH)(CH 3 ) 3 methylpropan-2-ol
(b)
The structural formula shown is that of trans 1,2-dibromobut-2-ene.
(i) Draw in the box provided, the corresponding cis isomer.
Br atoms both up or down
(ii) Discuss the conditions required for cis-trans isomerism to occur, using 1,2-dibromobut-2-ene as an example.
Cis-trans isomerism requires 2 conditions:
- A non-rotational C to C double bond which fixes the atoms in place
- Each double bonded C atom needs 2 different atoms/groups attached.
1,2-dibromobut-2-ene.has a C to C double bond. Each double bonded C atom is bonded to a Br atom and a CH3 group. Therefore this compound meets both requirements for cis-trans isomerism.
H 3 C C C CH 3
Br
Br
(ii) Oxidation to produce a carboxylic acid
Organic compound
Ethanol OR 1-hexene
Reagent
KMnO 4 acidified/heat
Organic product
Ethanoic acid OR hexan-1,2-diol
(iii) Addition
Organic compound
Hex- 1 - ene
Reagent
HCl
Major organic product
2 - chlorohexane
Minor organic product
1 - chlorohexane
(iv)Acid/base
Organic compound
Butanoic acid
Reagent
Na 2 CO 3
Major organic product
sodiumbutanoate
Other products
CO 2 and H 2 O
(c) When ethane and bromine react a substitution reaction takes place. When ethene and bromine react an addition reaction takes place. Use these two examples to discuss the differences between these two types of reaction. Your discussion should include: What is happening in the reaction The condition necessary for each reaction to occur The rate of the reactions The number of products of each reaction The structural formulae of the organic products of the reactions
Ethane + Br2 is a substitution reaction, where a Br atom replaces an H atom. This reaction requires UV light and time and is a slow reaction. 2 products are formed as shown:
CH 3 CH 3 + Br 2 → CH 3 CH 2 Br + HBr
Ethene + Br2 is an addition reaction, where the C to C double bond is broken and 1 new atom is added to each previously double bonded C atom to form a saturated compound. This reaction requires no special conditions, is quick and forms 1 product as shows:
CH 2 CH 2 + Br 2 →CH 2 BrCH 2 Br
QUESTION THREE
You are asked to identify three colourless liquids: pent-1-ene, butanoic acid and butan-1-ol. You are
provided with the reagents bromine water and acidified potassium dichromate.
In your answer you should:
Devise a scheme that will enable you to identify the three colourless liquids using only the reagents provided. Describe what you would do and the observations you would make. Write equations for any reactions described showing the organic reactants and products. You do not need to balance the equations.
Add Br 2 to a sample of each solution. In one sample, the orange Br 2 would decolourise rapidly, in the other
two it would remain orange. The one that decolourises quickly is the pent-1-ene – remove and label. To the
remaining compounds, add acidified dichromate. One sample would turn the orange dichromate green, the
other would remain orange. The one that changes to green is the butan-1-ol, the other butanoic acid.
Equation 1: CH 3 (CH 2 ) 2 CHCH 2 + Br 2 → CH 3 (CH 2 ) 2 CHBrCH 2 Br
Equation 2: CH 3 (CH 2 ) 3 OH + Cr 2 O 7 2-/H+^ → CH 3 (CH 2 ) 2 COOH
