KCSE 2024 CHEMISTRY PAPER2
Figure 1 shows part of the periodic table of elements. The letters are not the actual symbols of elements.
(a) Element Q belongs to period 5 and group VI. Place the element in the correct position in Figure 1. (1 mark)
(b) Consider the following ions: \(J^{+}\), \(K^{+}\) and \(M^{+}\).
(i) Write the electron arrangements for each. (2 marks)
I. \(J^{+}\) 2.8.8
II.\(K^{+}\) 2.8.8
III. \(M^{+}\) 2.8.8
3 correct = 2mks
2 correct = 1mk
1 correct = ½ mk.
(ii) Select the ion with the largest ionic radius. Give a reason. (2 marks)
\(J^{2-}\) It has the lowest nuclear charge / effective nuclear charge / lowest no of protons.
(c) Complete Table 1 by filling in the formula of the compound formed and the type of bond between the elements shown.
(d) Explain the following observations.
(i) Electrical conductivity of element F is higher than that of element E.
F has more delocalised electrons than E. / F has 3 delocalised electrons while E has 2 electrons.
(ii) Element M is a stronger reducing agent than element D.
M is more readily / easily loses electrons / M has a lower I.E / M is more electropositive / or the reverse
(iii) The melting point of element H is lower than that of element N.
Smaller size / H has smaller atomic size. So weak / less / fewer intermolecular forces / van-der-waals forces than N. / or (N opposite)
2. Figure 2 shows the steps in the contact process.

(a) Step 1 is known as the Frasch process. Describe how the process is carried out. (3 marks)
Superheated water (170°C) is pumped through the outermost pipe to the underground sulphur deposits to melt the sulphur.
Hot compressed air is then forced down the innermost pipe to push the molten sulphur to the surface through the middle pipe.
(b) State the optimum conditions used in step 3.
Catalyst: Vanadium (V) oxide \((V_{2}O_{5})\) or Platinum/ Platinized asbestos
Pressure: 2-3 atmospheres (or specific value 2-3 atm)
Temperature: 450-500°C (accept 400°C - 500°C) or specific temperature within 400°C - 500°C range
(c) Identify substance:
(i) A: Concentrated sulphuric (VI) acid \((H_{2}SO_{4})\)
(ii) B: Water \((H_{2}O)\)
(d) Name the process that occurs in step 2.
oxidation | roasting | combustion
(e)When concentrated sulphuric(VI) acid is added to glucose, a black solid is formed.
(i) Identify the black solid.
Carbon
(ii) State the property of concentrated sulphuric(VI) acid illustrated in this reaction.
Dehydrating agent | dehydration
3. The formulae of three organic compounds, each having two carbon atoms are;
Compound |
Formula |
A |
\(C₂H_{4}\) |
B |
\(C₂H_{2}\) |
C |
\(C₂H_{6}\) |
(a)State what is meant by the term homologous series.
Compounds with the same general formula but differ by a \(CH_{2}\) group, even if no mention of \(CH_{2}\).
(b)Compound B is the first member of its homologous series. Write the formula of the fifth member of the same series.
\(C_{6}H_{10}\)
(d) The flowchart in Figure 3 shows reactions involving compound B.

(i) The reagent used in
Step I is water (H₂O).
Step II: Hydrogen H₂
(ii) Identify the type of reaction that take place in :
Step IV: Hydration / Hydrolysis
(iii) cancelled
(iv) Draw the structure of compound Y.

(e) The following is a structure of a soap.
(i) Give the name of the main raw material used in making soaps.
Fat oil
(ii) Given two soaps, one with n = 16 and the other with n = 10, explain which one of the soaps is more effective in washing clothes.
Soap with n=16
It has a longer tail (carbon chain) which interacts with non-polar dirt (grease) more effectively. It has a longer hydrocarbon chain more hydrophobic.
4. Table 2 shows standard reduction potentials for given half cells.

a) (i) Draw a labelled diagram of an electrochemical cell using half cells II and III.

(ii) calculate the emf of the cell
\(E_{cell} = E_{red} - E_{oxid} \\ = -0.40 - (-1.66) \\ = +1.26 \text{ V}\)
(iii) Write the equation for the electrochemical cell
\(2Al + 3Cd^{2+} \rightarrow 2Al^{3+} + 3Cd\)
(b) Table 3 shows colours of aqueous ions.
Ions | Colour |
---|---|
Manganese(II) | Almost colourless |
Manganate(VII) | Purple |
Nickel(II) | Green |
State the observations made when a nickel rod is left standing in a beaker containing aqueous potassium manganate(VII). Explain.
Observation
The colour of the solution changes from purple to green
The nickel metal corrodes
Nickel dissolves (corrodes)
Explanation
Purple Manganate(VII) ions are reduced to colourless manganese (II) ions
Green nickel(II) ions are formed as nickel metal is oxidised and its ions released into the solution giving the solution a green colour. The corrosion of the nickel metal is as a result of formation of soluble nickel (II) ions
\(2\text{MnO}_4^- + 16\text{H}^+ + 5\text{Ni} \rightarrow 5\text{Ni}^{2+} + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}\)
(c) (i) One of the uses of electrolysis is electroplating. State one other use. (1 mark)
Extraction of reactive metals
Purification of metals
Synthesis / manufacture of chemicals e.g \(Cl_{2}\), NaOH, \(H_{2}\), \(O_{2}\)
Anodizing of aluminium
(ii) Silver is used to electroplate metals such as iron. State two properties of silver that make it suitable for this application. (2 marks)
It is attractive (shiny/beautiful/pleasant to see)
React slowly / Less reactive / Doesn't corrode
(iii) Figure 4 shows a set-up of an electrolytic cell used to electroplate an iron rod using silver.

Identify
R, T and U in Figure 4. (3 marks)R: Silver nitrate / \(AgNO_{3}\) / \(AgSO_{4}\) / \( Ag^{+} \)
T: Iron rod / Fe
U: Silver / Ag
(a) Explain how each of the following affects the rate of reaction:
(i) decrease in temperature:
Leads to decrease in the rate of reaction because the K.E. of the reactants is reduced leading to decreased number of effective/successful/fruitful collisions.
(ii) increase in surface area:
Leads to increase in the rate of reaction because there is increased contact area for more successful/fruitful/effective collisions to take place.
(b) Using a 250 ml volumetric flask, a burette and 12.0 M hydrochloric acid, describe how a standard solution containing \(250 cm^{3}\) of 0.5 M hydrochloric acid can be prepared. (3 marks)
C1V1 = C2V2 ⇒ V1 = C2V2 / C1 = (0.5 x 250 )/ 12 = 10.42 \(cm^{3}\)
Accept 10.4, 10.41, 10.416 \(cm^{3}\)
- Using a burette, place 10.42 cm3 of 12.0 M HCl into a 250 \(cm^{3}\) volumetric flask (only 10.42)
- Top up to the 250 ml mark with distilled water
- The solution formed is 0.5 M Hcl
(c) 5.0 g of zinc powder was reacted with 25.0 cm³ of 0.5 M hydrochloric acid. The volume of gas produced was measured every 10 seconds. Table 4 shows the data obtained.
Time (seconds) | Volume of hydrogen gas (cm3) |
---|---|
0 | 0 |
10 | 52 |
20 | 86 |
30 | 110 |
40 | 128 |
50 | 136 |
60 | 140 |
70 | 140 |
80 | 140 |

(ii) From the graph, determine the rate of reaction at:
I) 5 seconds
II) 37 seconds
Note:
Time (s)
Tangent at 5 seconds
Rate of reaction = change in concentration / change in time
Scale must be more than half on both axes, labeled with correct units
-ve graph inverted, max 2mks
(iii) Give a reason for the difference in the rates calculated in (c)(ii) I and II. (1 mark)
value in (c) ii (I) must be greater than of (c) ii (II)
explanation/reason: Rate of reaction decreases with time because the reactants are being used up as reaction progresses.
(iv) State one observation that would be made if the experiment was repeated using 5.0 g of zinc powder and 25.0 cm³ of 0.25 M hydrochloric acid. (1 mark)
The rate of evolution of hydrogen gas is less. Total volume of hydrogen gas produced is half. Less effervescence.
6. (a)State the meaning of the term standard molar heat of combustion? (1 mark)
Enthalpy Change when one mole of a substance burns completely in air at standard condition
(b) Table 5 gives the standard enthalpies for three reactionsReaction | Equation | ΔH (kJmol-1) |
---|---|---|
I | H2(g) + ½ O2(g) → H2O(l) | -286 |
II | Si(s) + O2(g) → SiO2(s) | -911 |
III | SiH4(g) + 2O2(g) → SiO2(s) + 2H2O(l) | -1517 |
The chemical equation shows the reaction between silicon and hydrogen to form silane:
Si(s) + 2H₂(g) → SiH₄(g)
Calculate the enthalpy change for this reaction using the information in Table 5. (3 marks)

c) Determine the amount of energy change when 1 kg of water is formed. (1 mark)
(H = 1.0, O = 16.0)
RFM of H₂O = 18g
18g → -286 kJ
1000g → ?
= -15888.89 kJ = -15889 kJ
(d) Heating value of a fuel is the amount of heat energy released when 1 g of a substance undergoes combustion.
Calculate the heating value of carbon and hydrogen using the following information.
C(s) + O₂(g) → CO₂(g); ΔH = -393 kJmol⁻¹
H₂(g) + 1/2 O₂(g) → H₂O(l); ΔH = -286 kJmol⁻¹
(C = 12.0; H = 1.0; O = 16.0).
(i) Carbon. (1 mark)
2g = -393 kJ
1g = ?
1g x -393/12 = -32.75 kJ/g
Ignore the sign
Heating value of Carbon = 32.75 kJ/g
(ii) Hydrogen
Calculation for Hydrogen:
2g = -286 kJ
1g = ?
1 x 286 kJ / 2 = -143 kJ/g
Heating value of Hydrogen = 143 kJ/g
(e) Metals V, W and X displace copper from its compounds. Describe an experiment that can be carried out to arrange the three metals in order of their reactivity with copper using aqueous copper(II) sulphate and a thermometer. (3 marks)
Place equal volumes of aqueous CuSO₄ in three beakers.
To each beaker, add excess of each metal powder.
Determine the temperature change and arrange the metals in order of reactivity.
The reaction with the highest temperature change has the most reactive metal.
7 (a) There are two types of water hardness. One type is permanent hardness caused by the presence of calcium and magnesium ions.
(i) I. Give the name of the other type of water hardness.
Temporary
(ii) II. Name the ion responsible for the type of water hardness named above.
HCO₃⁻ (Hydrogen Carbonate ion)
(iii) State one natural source of calcium ions in river water.
Dissolution of calcium containing rocks (limestone, gypsum) by acidic soil and rain.
iii) Describe how ion exchange can be used to remove permanent hardness in water.
Ion exchange uses a resin containing either Sodium or Hydrogen ions that exchange with \(Mg^{2+}\) and \(Ca^{2+}\) ions. Hard water is passed through the resin and Ca2+ ions are replaced with Sodium or Hydrogen. Water comes out as soft water
(b) Figure 5 shows solubility curves of K₂O and KCl in grams per 100 g of water.

(i) State the temperature at which the solubility of potassium chloride is the same as that of potassium nitrate. (1 mark)
20°C ± 1°C
16 (ii) A solution that contains 40 g of potassium chloride in 100 g of H₂O is cooled slowly from 75°C.
I. State the temperature at which crystals start to form. (1 mark)
40°C ± 1°C
II. Determine the mass of potassium chloride that will form if the solution is cooled to 15°C. (1 mark)
40.0 - 32.5 = 7.5g ± 0.5g
(iii) 40 g of potassium nitrate is dissolved in 50 g of water at room temperature. If the mixture is slowly heated, determine the lowest temperature at which all the solid dissolves. (2 marks)
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