VCE Chemistry Unit 3 Area of Study 2

^{(Chapter 6 rates of reactions)}

Which of the following changes will definitely lead to a faster reaction rate?

• lower temperature, addition of a catalyst and higher concentrations

• higher temperature, lower concentrations and the addition of a catalyst

• greater surface area, higher temperatures and higher concentrations

• greater surface area, lower temperatures and lower concentrations

^{(Chapter 6 rates of reactions)}

A series of reactions are conducted to compare the reaction rates between calcium carbonate and hydrochloric acid. The time taken for each reaction to produce 80 mL of gas is recorded.

Which of the following is unlikely to be a useful method for monitoring the rate of this reaction?

• Monitoring the colour change occurring during the reaction.

• Measuring the volume of gas evolved.

• Monitoring the mass of the flask.

• Monitoring the pH of the solution.

^{(Chapter 6 rates of reactions)}

A series of reactions are conducted to compare the reaction rates between calcium carbonate and hydrochloric acid. The time taken for each reaction to produce 80 mL of gas is recorded.

In a comparison of experiment 2 and experiment 3, it is likely that

• the temperature in experiment 3 was higher than that of experiment 2.

• the calcium carbonate particle size in experiment 3 is larger than experiment 2

• the rate of reaction in experiment 3 was greater due to the increased calcium carbonate mass.

• the rate of reaction in both reactions should have been the same as the HCl concentration was the same

^{(Chapter 7 Extent of reactions)}

The equilibrium expression shown refers to which equation?

• 
• 
• 
• 

^{(Chapter 7 Extent of reactions)}

The reaction between sulfur dioxide and oxygen is

The numerical value of K for this reaction at 100°C is 12.4

The concentrations in a mixture of the above gases at 100°C, are

[SO₂] = 0.4M [O₂] = 0.4M [SO₃] = 0.4M

• This reaction mixture is at equilibrium.

• The forward reaction needs to occur more than the back reaction to reach equilibrium.

• The back reaction needs to occur more than the forward reaction to reach equilibrium.

• The back reaction will stop until equilibrium is reached.

^{(Chapter 7 Extent of reactions)}

The reaction between sulfur dioxide and oxygen is

The numerical value of K for this reaction at 100°C is 12.4

4.8 mole of SO₂ and 3.0 mole of O₂ is introduced to an empty reactor. When equilibrium is reached

• 4.8 mole of SO₃ will have formed

• 6.0 mole of SO₃ will have formed.

• SO₂ will not be present as it is the scarce reagent.

• The amount of SO₃ formed will equal the amount of SO₂ reacted.

^{(Chapter 7 Extent of reactions)}

The reaction between sulfur dioxide and oxygen is

The numerical value of K for this reaction at 100°C is 12.4

The correct unit for the equilibrium constant in this reaction will be

• M^-1

• M¹

• M²

• no unit at all since the number of product molecules will equal the reactant molecules.

^{(Chapter 7 Extent of reactions)}

Hydrogen and iodine can react in a reversible reaction:

H_2(g) + I_2(g) ↔ 2HI_(g)

If the value of K for this reaction at 50°C is 34.4, what is the value of K at the same temperature for

• 0.029

• 0.170

• 17.2

• 34.4

^{(Chapter 7 Extent of reactions)}

In the decomposition of NOCl, the reaction is

In an equilibrium mixture at 30°C, the amount of

• Cl₂ is half the amount of NOCl

• NO will equal the amount of NOCl

• NO is far less than the amount of NOCl

• NO added to the amount of Cl₂ will give the amount of NOCl.

^{(Chapter 7 Extent of reactions)}

A catalyst is added to an equilibrium mixture of gases. The catalyst will

• increase the rate of the forward reaction, changing the value of K.

• increase the rate of the forward reaction but K is unchanged.

• increase the rate of both the forward and back reactions.

• have no impact on the mixture.