Questions
- 6g of Potassium nitrate solid was added to 120cm3 of water in a plastic beaker.
The mixture was stirred gently and the following results were obtained.
Initial temperature = 21.5oC
Final temperature = 17.0oC- Calculate the enthalpy change for the reaction (Density =1g/cm3, C= 4.2 Jg-1K-1)
- Calculate the molar enthalpy change for the dissolution of potassium nitrate (K=39, N= 14, O =16)
- The heat of combustion of ethanol, C2H5OH is 1370 KJ/mole.
- What is meant by heat of combustion?
- Calculate the heating value of ethanol (H = 1.0, C = 12.0, O = 16.0)
- The heat of combustion of ethanol, C2H5OH is 1370 KJ/mole.
- Use the information below to answer the questions that follow:-
Ca(s) + ½O2(g) → CaO(s) ΔH =-635KJ/mol
C(s) + O2(g) → CO2(g) ΔH= -394KJ/mol
Ca(s) + C(s) + 3/2O2(g) → CaCO3 ΔH = -1207KJ/mol
Calculate the enthalpy change for the reaction:
Ca(s) + CO2(g) → CaCO3(s) - 0.92g of ethanol were found to burn in excess air producing a temperature rise of 32.5ºC in 200cm3 of water.
C=12.0 H=1.0 O=16.0
Density of water 1g/cm3
Specific heat capacity of water 4.2 KJ kg-1K-1- Write the equation for combustion of ethanol
- Determine the molar heat of combustion of ethanol
- Study the information in the following table and answer the questions that follow. The letters do not represent the actual chemical symbols of the elements.
ELEMENT U V W X Y Z NUMBER OF PROTONS 18 20 6 16 19 17 NUMBER OF NEUTRONS 22 20 8 16 20 20
Which of the above elements are:- Likely to be radioactive?
- Able to form a compound with the highest ionic character?
- The diagram below shows energy levels for the reaction
½ H2(g) + ½F2(g) → HF(g)- Work out the activation energy for the reaction
- Calculate the heat of formation of HF
- Is the reaction endothermic or exothermic?
- Using the heats of combustion of the following substances, calculate the heat of formation of ethanol
C(s) + O2 (g) →CO2 (g); ΔH = -393KJmol-1
H2 (g) + ½ O2 (g) → H2O(l); ΔH = -286KJmol-1
CH3CH2OH(l) + O2 (g) → 2CO2 (g) + 3H2O (l) ;ΔH = 1386KJmol-1 - Nitrogen and hydrogen react reversibly according to the equation:-
N2(g) + 3H2(g) → 2NH3(g); ΔH = -92kjmol-1
The energy level diagram for the above reaction is shown below:-- How would the yield of ammonia be affected by:
- A decrease in temperature
- An increase in pressure
- How does a catalyst affect reversible reaction already in equilibrium?
- On the above diagram, sketch the energy level diagram that would be obtained when iron catalyst is added to the reaction
- How would the yield of ammonia be affected by:
- Study the table below and answer the questions that follow
Bond type bond energy kJmol-1
C - C 346
C = C 610
C - H 413
C - Br 280
Br - Br 193- Calculate the enthalpy change for the following reaction
C2H4(g) + Br2(g) → C2H4Br2(g) - Name the type of reaction that took place in (a) above
- Calculate the enthalpy change for the following reaction
- Bond energies for some bonds are tabulated below:-
BOND BOND ENERGY KJ/mol H – H 436 C = C 610 C- H 410 C - C 345
C2H4(g) + H2(g) → C2H6(g) - The able shows the results obtained when 20.2g of potassium nitrate was added in 50cm3 of water.
Time in (min) 0.0 0.3 1.0 1.3 2.0 2.3 3.0 3.3 4.0 Temperature (oC ) 25.0 25.0 25.0 25.0 17.0 17.0 20.0 20.0 20.0 - Draw the graph of temperature against time
- Using the graph, determine the temperature change
- Calculate the heat change
- Find the molar heat of solution of potassium nitrate
- When 1.6g of ammonium nitrate were dissolved in 100cm3 of water, the temperature dropped by 6ºC. Calculate its enthalpy change. (Density of water = 1g/cm3, specific heat capacity is 4.2kJ kg-1K-1)
- Sodium hydrogen carbonate was strongly heated.
- Write an equation for the reaction
- The grid below shows part of the periodic table. Use it to answer the questions that follow. The letters are not the actual symbols.
- Write the equation for the reaction that occurs between elements L and D
- The oxide of G reacts with both hydrochloric acid and sodium hydroxide. What is the nature of the oxide of G?
- Explain why elements H has a higher boiling points than element D.
- State one use of element E
- Compare and explain the atomic radius of B and C
- 11.5g of L was completely burnt in oxygen .Calculate the volume of gas that was used. (L = 23, molar gas volume at room temperature is 24dm3)
- A student has been provided with sodium hydroxide solution of 2M and hydrobromic acid of 4M. He was asked to investigate the equation for the reaction between these two substances and hence determine the molar enthalpy of neutralization. He carried out the reaction and obtained the following results:-
Vol. of 4M Hydrobromic acid added to 20cm3 of 2M NaOH Temperature of the mixture (oC) 4.0 26.8 6.0 30.0 8.0 33.2 10.0 36.0 12.0 35.2 14.0 34.4 20.0 30.8 - Draw a graph of the temperature of the mixture (vertical axis against the volume of the acid added)
- Using the graph estimate the temperature of the mixture when 17cm3 of the acid was added
- Both solutions were at room temperature at the start of the experiment. Use your graph to estimate the room temperature
- What is the significance of the highest temperature of the solution mixture?
- The temperature of the mixture increased during the first additions of the acid. Why did the temperature increase?
- Suggest a reason why the temperature decreased during the latter part of the experiment
- Use your graph to determine the volume of 4M Hydrobromic acid which just neutralize 20cm3 of 2M NaOH
- How many moles of Hydrobromic acid are present in your answer in (g) above?
- How many moles of NaOH are present in 20cm3 of 2M of NaOH solution?
- Use your answers in (h) and (i) above to write an equation of the reaction taking place in the experiment. Explain clearly how you have used your answers (1½mks)
- Determine the molar enthalpy of neutralization of hydrobromic acid (1½mks)
-
- The following results were obtained in an experiment to determine the enthalpy of solution of sodium hydroxide
Mass of plastic beaker = 8.0g
Mass of plastic beaker + distilled water = 108.15g
Mass of plastic beaker + distilled water + sodium hydroxide = 114.35g
The table below shows the temperature at fixed times after mixing
Time/seconds 0 30 60 90 120 150 180 210 Temperature ( oC) 15 21 29 28 27 26 26 25 - Plot a graph of temperature (y-axis) against time (x-axis)
- From your graph, determine the maximum temperature attained
- Determine the temperature change of the reaction
- Calculate the number of moles of sodium hydroxide used in the experiment
(Na = 11, H = 1, O = 16) - Use your results to determine the molar enthalpy solution of sodium hydroxide. (Density of solution is 1g cm-3 , specific heat capacity of solution = 4.18 KJ-1K-1)
- Examine the energy level diagram below and use it to answer the questions that follow
- Which ΔH values will have negative sign?
- What physical change is being represented where enthalpy change ΔH4 is involved? ( ½mk)
- In terms of ΔH1, ΔH2, ΔH3 and ΔH4, give the overall enthalpy change for the reaction:-
H2(g) + ½O2(g) → H2O(l) - Is the reaction in (iii) above exothermic or endothermic?
- The following results were obtained in an experiment to determine the enthalpy of solution of sodium hydroxide
-
- Distinguish between molar latent heat of fusion and molar latent heat of vaporization
- Study the graph below and answer the questions which follow:
- Explain the changes occurring between points
BC ………………………………………… CD ……………………… - In an experiment to determine molar enthalpy of neutralization of hydrochloric acid using potassium hydroxide, the data below was obtained. The concentration of potassium hydroxide used was 0.5M
Volume of 0.5M KOH (cm3) 0 5 10 15 20 25 30 35 Total volume of acid + Base 20 25 30 35 40 45 50 55 Temperature (oC) 24 26 27 28 29 29 28 27 - Plot a graph of temperature (y-axis) against volume of potassium hydroxide used
- From your graph:
- Determine the temperature change
- Find the volume of potassium hydroxide which completely neutralized 20cm3 of the acid
- Calculate the heat change for the reaction (C = 4.2Jg-1K-1 density of solution = 1g/dm3)
- Calculate the molar enthalpy of neutralization of hydrochloric acid with potassium hydroxide
- Explain the changes occurring between points
- A typical electrolysis cell uses a current of 40,000 amperes. Calculate the mass (in Kg of aluminium produced in one hour). (Al = 27) (Faraday = 96500 Coloumbs)
- Biogas is a mixture of mainly Carbon (IV) Oxide and methane.
- Give a reason why biogas can be used as a fuel
- Other than fractional distillation, describe a method that can be used to determine the percentage of methane in biogas
- Consider the following equilibrium reaction.
H2(g) + Cl2(g) ⇌ 2HCl(g) ΔH= -74.4KJ
State and explain the effect of formation of hydrogen chloride if pressure was increased in the equation above - Turning of fossil fuels has adverse environmental effects:-
- Name two pollutants from the burning of petroleum products
- Give one precaution taken to minimise the pollution by fossil fuels
-
- Define molar heat of neutralization
- The rise in temperature when 50cm3 of sodium hydroxide is reacted with two acids is given in the table below:-
Acid 50cm3 of HCl 50cm of Oxalic acid Temp rise (oC) 7 4
Explain the difference in the temperature.
- Calculate the latent heat of vaporization of water
H2O(l) → H2O(g)
Given the following thermo chemical equations:-
H2(g) + ½O2(g) → H2O(g) ΔHθ= -242KJ/Mol
H2(g) + ½O2(g) → H2O(l) ΔHθ= -286KJ/Mol - Define the term fuel
- State four reasons why wood fuel is chosen for domestic cooking
- Define the term fuel
- The setup bellow was used to investigate the changes that take place when sodium hydroxide pellets dissolve in water.
- Why is a plastic beaker used instead of a metallic beaker?
- State and explain the observations made in the above reaction
- What is a fuel? (1mark)
- Other than the cost, state two other factors to consider when choosing a fuel.
- The equation below represents changes in the physical state of ions metal:
Fe(s) → Fe(l) ΔH= +15.4kjmol-1
Fe(l) → Fe(g) ΔH=+354kjmol-1- Calculate the amount of heat energy required to change 10kg of solid iron to gaseous iron (Fe = 56)
- Iodine can react with chlorine as shown below:-
I2(g) + Cl(g) → 2lC(s) ΔH= -68kJ
Determine the molar enthalpy change for this reaction - Draw an energy level diagram for the reaction in (b) above
- Study the diagram below and answer the questions that follow:
- What do ∆H1 and ∆H2 represent?
∆H1 …………………………………………………………………….
∆H2 …………………………………………………………………….. - Write an expression to show the relationship between ∆H1, ∆H2 and ∆H3
- What do ∆H1 and ∆H2 represent?
Answers
-
- ΔH = 120 x 4.2 x 4.5 ( ½mk)
1000
= + 2.268KJ (½mk) - RFM of KNO3 = 39 + 14 + 48 = 101
6g → 2.268KJ
101g → ?
101 × 2.268 (½mk)
6
= + 38.178KJ mol-1 (½mk
- ΔH = 120 x 4.2 x 4.5 ( ½mk)
-
- Heat evolved when one mole of a substance is completely burnt in oxygen
- RFM of C2H5OH = 46
Molar mass = 46g
Heating value = 1370 KJ
46g
= 29.78KJ/g (with units)
- Ca(s) + C(s) + 3/2O2 (g)
-
- C2H6O(l) + 3O(g) → 2CO2(g) + 3H2O
- ΔH = MCΔT
200 × 4.2 × 32.5 = -27.3Kj
1000
0.92g C2H6O = - 27.3Kj
46g C2H6O = ?
46g × 27.3Kj = -1365Kj
0.92
ΔHC C2H6O = -1365Kj mol
-
- U,V,Y,Z All the 4 or nay 3 exclusively correct penalize ½ mk if wrong answer
- YZ is/are included any 2 correct ½ mk
-
- 611 − 389 = +222 KJ
- H = +222 – (611 – 100)
= -289KJ - Exothermic reaction
-
∆Hf + ∆H3 = ∆H1 + ∆H2
∴ ∆Hf = ∆H1 + ∆H2 - ∆H3 √½
= -393 x 2 + -286 x 3 +1386 √1
= -786 – 858 + 1386
= -1644 + 1386 √1
∆Hf = -258 KJmol-1 √½ -
-
- the yield of NH3 would be lowered √ ½ any supply of heat makes NH3 to decompose to N2 and H2
- the yield of NH3 would be increased
- a catalyst accelerate the rates of both forward and reverse reactions equally√ ½ . Equilibrium position is not affected by a catalyst√ ½
-
-
-
- Breaking of ‘C = C’ = +610 KJ
Breaking of ‘Br – Br’ = +193
803√
Formation of 2C – Br = -560
Formation of C - C +243 Kj
-346
- 103KJ√ 2 marks - Addition reaction/ halogenation √
- Breaking of ‘C = C’ = +610 KJ
Bond breaking Bond formation
4 C-H – 4 x 410 = 1640 6C – H 6 x 410
C = C – 1 x 610 = 610 = 2460
H – H – 1 x 436 = 436 C – C – 345
2686 2805
ΔH = 2686 – 2805
= -119 Kj/Mol-
- Graph
labeling -*TZM*
plotting – *TZM*
scale – *TZM*
line – *TZM*
total 5mks - Shown on the graph -*TZM*
- Heat change = MCT
= 50 x 4.2 x 10.2
100
= 2.142kJ - RFM of KNO3 = 39 + 14 + 48
= 101
H = 2.142 x 101 = -10.71Kjmol-1
20.2
- Graph
- MCT = 100 × 4.2 × 6 = 2.52 Kj
1000
Moles of NH4NO3 = 1.6 = 0.02 moles
80
If 0.02 mol = 2.52 Kj
1 mol = ?
1 × 2.52 = +126KJ/ mol
0.02 -
- 2 NaHCO3 (g) → Na2CO3(g) + H2O(l) + CO2(g)
-
- 2L(g) + D2(g)→ 2LD(g)
- Amphoteric oxide
- Element H has a giant atomic structure with strong covalent bonds throughout its structure while D has simple molecular structure with weak Vander wall forces (2 m)
- - Used in advertising signs (Advertisements)
- Used in florescent tubes (Any two correct use) - C has a smaller atomic radius than B because it has stronger nuclear charge// more number of protons which attract the outer energy level electrons more firmly (2 mks)
- 4L(s) + O2(g) → 2 L2O(g)
Moles of L = 11.5/23 = 0.5 moles
Moles of O2 = 0.5/0.4 = 0.125 moles
Volume of O2 = 0.125 mol X 24 = 3 dm3
4L (s) + O2(g) → 2L2O(s)
If 4 x 23g = 24dm3
11.5g of L = ?
11.5 x 24 = 3dm3
4 x 23
-
- Drawn on the graph
A = ½ mk
S = ½ mk
P = ½ mk
C = ½ mk - 32.5oC + 1 Read from the student’s correctly plotted graph.
- 20oC + 0.5 Line is extrapolated downwards from the student’s correct graph.
- It is end point/ complete neutralization.
- The reaction is exothermic hence as reaction proceeded more heat was produced.
- Reaction was complete hence solution lost heat through radiation to the surrounding.
- 10.2 cm3 + 0.1. Read from the student’s correct graph.
- Moles = M x V
1000
= 10.2 x 4 √ ½ = 0.0408 moles √ ½
1000 - Moles = M x V
1000
= 2 x 20 √ ½ = 0.04 moles √ ½
1000 - HBr : NaOH
0.0408 : 0.04
0.0408 : 0.04
0.04 0.04
1 : 1
HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l) - ∆H = MC∆t
= -30.2g x 4.2J x 16.3
g0c
= -2067.49J √ ½
Ans. in (h) = -2067.49 J.
∴ 1 Mole = 1 x 2067.49J √ ½ e.g. 1 x 2067.49
Ans in “h” 0.0408
= -Ans. e.g 50673.82 J mol-1
Or 50.67382KJ mol-1 √ ½
- Drawn on the graph
-
-
- Max. temperature attained : 290c
- Temperature change of the reaction = (29-15)0c
= 140c - Mass of NaOH used = (114.35 – 108 .15)g
= 6.2g
R.F.M of NaOH = 40g
Moles of NaOH used = 6.2/40 moles
= 0.155moles - Heat released = Mass × Specific Heat Capacity × Temperature change
Mass of water used = (108.15 – 8)g
= 100.15g
∴ Heat released = 100.15/1000 × 4.18 × 14 kj
=100.15kj
0.155 moles NaOH → 5.861 kj
1 mole NaOH → ?
1 x 5.861 kjmol-1
0.155
= -37.8 kjmol-1
-
- ΔH3 and ΔH4
- Condensation
- ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
- Exothermic.
-
-
- – Latent heat of fusion is the heat change that occurs when one mole of a solid substance changes into liquid at constant temperature.
- Latent heat of vapourization is the heat change that occurs when one mole of liquid substance changes into gas at constant temperature. -
- – BC – The liquid loses heat as it cools hence decrease in kinetic energy of the particles
- CD - The liquid changes to solid as temperature remains constant at freezing point. -
- Scale – *TZM*
Plot – *TZM*
Line -
- Should be shown on the graph – if not shown penalize ( ½ mk)
- Scale – *TZM*
- Heat change = m x c x ΔT
Where m = (vol. of acid (20cm3) + volume of bas in (b) above) x 1g/cm3
ΔT-as read form the graph - moles of acid
Moles of base = 0.5 x volume in (b) above
1000
Mole ratio acid: Base = 1:1
Moles of acid → heat change in (iii)above
1mole → ?
Molar heat change = 1 x heat in (iii)
Moles of acid
- – BC – The liquid loses heat as it cools hence decrease in kinetic energy of the particles
- – Latent heat of fusion is the heat change that occurs when one mole of a solid substance changes into liquid at constant temperature.
- Q = 40000 x 60 x 60 = 144000000c
Mass of Al = 144000000 x 27
3 x 96500
= 13.43kg -
-
- Contains methane which is a fuel or contains methane which can burn
- Pass a known volume of biogas through Sodium hydroxide (Potassium hydroxide) solution to absorb Carbon (IV) Oxide. Measure the volume of remaining gas
% = Volume of methane x 100
Volume of Biogas
-
- No effect – Reaction is not accompanied by volume changes/ similar volumes of reactants and products
-
- – carbon IV Oxide;
- Sulphur IV Oxide;
- Lead; - Availed low sulphur diesel/ availed unleaded petrol
- – carbon IV Oxide;
-
- Heat change that occurs when one mole of hydrogen combines with one mole of hydroxide ions. //Heat evolved when one mole of water s formed during reaction of H+ and OH- ions
- HCl produces a higher temperature rise than oxalic acid;
HCl is a stronger acid than oxalic acid;
-
ΔH2 = - ΔH1 + ΔH3
= ΔH3 – ΔH1
= -242 - -286
= -242 + 286
= +44KJ/mol (No units of sign = ½mk) -
- Chemical substance that burns to produce useful amount of heat.
- - Its cheap
- Its readily available (½mk)
- It burns slowly (½mk)
- Does not produce poisonous gas. (½mk)
-
- Metallic beaker would make most of the heat be lost to the environment
- - Thermometer reading increased
- The reaction is exothermic
-
- A substance that produce heat energy when burnt
- - Availability
- ease of transport
-
- 1 mole Fe (56) required → 15.4 + 354 = 396.5Kj
10,000 (10 kg) → ?
10,000g × 369.5 Kj
56g
= 6596.285Kj - - 68Kj/2 = - 34 Kj ✓½
- 1 mole Fe (56) required → 15.4 + 354 = 396.5Kj
-
- ∆H1 – Lattice energy ✓1
∆H2 – Hydrogen energy ✓1 - ∆H3 = ∆H2 + ∆H1 ✓1
- ∆H1 – Lattice energy ✓1
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