Test Thermochemistry Form 5

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 28^oC to 48^oC. Given that the specific heat capacity of the solution is 4.2 J g^-11^oC^-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^-11^oC^-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

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-4