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WorksheetsTest Thermochemistry Form 5
Total questions: 15
Worksheet time: 42mins
A pupil carried out an experiment to determine the value of heat of displacement. Figure 3 shows the set up of the apparatus used in the experiment. The following data was obtained:
Initial temperature of copper(II) sulphate solution, θ1, = 28 °C
Highest temperature of the mixture of products, θ2 = 48 °C
Complete the ionic equation for the reaction that occurred.
CuSO4 + Zn2+ → ZnSO4 +Cu2+
Zn2+ + Cu → Zn + Cu2+
CuSO4 + Zn → ZnSO4 +Cu
Zn + Cu2+ → Zn2+ + Cu
A pupil carried out an experiment to determine the value of heat of displacement. Figure 3 shows the set up of the apparatus used in the experiment.
In this experiment, excess zinc is added to 100 cm3 of 0.5 mol dm-3 copper(II) sulphate solution. The temperature of the mixture rise from 28oC to 48oC. Given that the specific heat capacity of the solution is 4.2 J g-11oC-1 and the density of the solution is 1.0 g cm -3
Calculate the heat release in the experiment.Use the formula, ΔH = mcθ
4200 J
8500 J
8400 J
4300 J
The thermochemical equation for the neutralisation reaction between nitric acid and sodium hydroxide solution is given below.
HNO3 + NaOH → NaNO3 + H2O , ΔH = -57.3 kJ/mol
State the meaning of heat of neutralisation based on this experiment?
Heat released when 1 mole of water is formed from the reaction between acid and alkali
57.3 kJ/mol of heat released when 1 mole of water is formed from the reaction between nitric acid and sodium hydroxide
Heat released when 1 mole of water is formed from the reaction between nitric acid and sodium hydroxide
57.3 kJ/mol of heat released when 1 mole of hydroxide ion is formed from the reaction between nitric acid and sodium hydroxide
Based on the given thermochemical equation below, state one observation when dilute nitric acid is added to sodium hydroxide solution.
HNO3 + NaOH → NaNO3 + H2O , ΔH = -57.3 kJ
Choose two correct answers:
The beaker/container become hot
The beaker/container become cold
Thermometer reading increase
Yellow solution formed
In an experiment, 100 cm3 of 2 mol dm-3 nitric acid solution was added to 100 cm3 of 2 mol dm-3 sodium hydroxide solution.
[Specific heat capacity of solution = 4.2 Jg-1°C-1; Density of solution = 1 g cm-3]
The thermochemical equation for the neutralisation reaction is given below.
HNO3 + NaOH → NaNO3 + H2O , ΔH = -57.3 kJ
Calculate the heat energy released in this experiment.
1.146 J
1.146 KJ
1146 KJ
11.46 KJ
In an experiment, 100 cm3 of 2 mol dm-3 nitric acid solution was added to 100 cm3 of 2 mol dm-3 sodium hydroxide solution.
Draw the energy level diagram for the reaction between nitric acid and sodium hydroxide.
In an experiment, 100 cm3 of 1.0 mol dm-3 of sodium hydroxide solution is added to 100 cm3 of 1.0 mol dm-3 hydrochloric acid. The heat of neutralization obtain is -57.3
kJ/mol
Calculate the temperature change in the experiment.
[Specific heat capacity of solution = 4-2 Jg-1 °C, Density of solution =1 g cm-3]
6.80 °C
6.82 °C
13.64 °C
13.60 °C
Table above shows the values of heat of neutralisation, ΔH for a reaction of sodium hydroxide solution with two different acids.
Compare AND explain why there is a difference in the values of heat of neutralisation.
The heat of neutralisation between NaOH and HCl is higher
The heat of neutralisation between NaOH and CH3COOH is lower
HCl is a strong acid and ionise completely in water to produce H+ ion, while CH3COOH is a weak acid and ionise partially in water to produce H+ ion
Some heat given out is used to dissociate ionises CH3COOH /molecules completely.
Excess magnesium is added to 100 cm3 of 0.5 mol dm-3 copper(II) sulphate solution. Given that the specific heat capacity of the solution is 4.2 J g-11oC-1 and the density of the solution is 1.0 g cm -3, calculate the heat of displacement in the experiment. The results obtained are shown below:
Initial temperature of copper(II) sulphate solution, θ1, = 25 °C
Highest temperature of the mixture of products, θ2 = 45 °C
-168 kJ mol-1
-168 J mol-1
168 kJ mol-1
-1680 J mol-1
During an exothermic reaction, heat content in the surroundings increases because
heat energy is destroyed during reactions
the reaction absorbs heat energy
the energy contained in the reactants is lower then the products
the energy contained in the products is lower than the reactants
