Unit notes and practice questions
Everything here is for unit 4 of the Grade 11 Chemistry textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.
The cards filed under this chapter on the Grade 11 Chemistry page. Open one at a time.
When you use it
Use to explain why collisions between reacting species succeed or fail to form products.
Watch out
A reaction requires both conditions simultaneously: reactant molecules must collide with proper orientation and with kinetic energy equal to or greater than the activation energy.
Drafted from Grade 11 Chemistry, pages 193-221, then checked twice before it went up
When you use it
Use this to relate the rate of reaction to the consumption of reactants or the formation of products for any balanced chemical equation.
Watch out
Always divide the rate of concentration change of each species by its stoichiometric coefficient in the balanced equation.
Drafted from Grade 11 Chemistry, pages 193-221, then checked twice before it went up
When you use it
Use when writing rate expressions in terms of disappearing reactants.
Watch out
Reactant concentration decreases over time, making change in concentration negative. You must include a minus sign in front of reactant terms because rate of reaction is always positive.
Drafted from Grade 11 Chemistry, pages 193-221, then checked twice before it went up
10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.
Question 1
The rate of a chemical reaction is best defined as:
Answer: B the change in concentration of a reactant or product per unit time
Why
The chapter says the rate of a chemical reaction measures the change in concentration of a reactant or a product per unit time. Mass at the end, starting temperature and amount of catalyst are not the definition.
Question 2
Why does increasing the temperature of a reaction mixture usually increase the rate of reaction?
Answer: A The average kinetic energy of the reacting particles increases, so they collide more frequently and with greater energy.
Why
The chapter explains that when temperature rises, the average kinetic energy of the particles increases. This makes the particles collide more often and with more energy, so the rate goes up.
Question 3
Why do finely divided solids react more rapidly than the same mass of the same substance in a large lump?
Answer: C The powder has a greater surface area available for the reaction.
Why
The chapter says finely divided solids react more rapidly because of the greater surface area available. More surface means more particles are exposed to the other reactant at one time.
Question 4
Which statement about a positive catalyst is correct?
Answer: C It lowers the activation energy, so the reaction becomes faster.
Why
The chapter says a positive catalyst increases the rate of reaction by lowering the activation energy, Ea. Catalysts are not used up and do not change the position of equilibrium.
Question 5
For the reaction 3O2(g) → 2O3(g), which rate expression is correct?
Answer: B Rate = -(1/3)Δ[O<sub>2</sub>]/Δt = (1/2)Δ[O<sub>3</sub>]/Δt
Why
The coefficients are 3 for O2 and 2 for O3. Each reactant term is divided by its coefficient (1/3 for O2), and each product term is divided by its coefficient (1/2 for O3). A minus sign is used for the reactant because its concentration decreases.
Question 6
In the reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g), the rate of disappearance of NO is 4.0 × 10-5 mol L-1 s-1. What is the rate of formation of N2?
Answer: C 2.0 × 10<sup>-5</sup> mol L<sup>-1</sup> s<sup>-1</sup>
Why
NO has a coefficient of 2 and N2 has a coefficient of 1. So the rate of N2 formation equals (1/2) of the rate of NO disappearance: (1/2) × 4.0 × 10-5 = 2.0 × 10-5 mol L-1 s-1.
Question 7
For the decomposition 2N2O5(g) → 4NO2(g) + O2(g), the average rate of formation of NO2 between 300 s and 400 s is 4.8 × 10-5 mol L-1 s-1. What is the average rate of formation of O2 in the same period?
Answer: D 1.2 × 10<sup>-5</sup> mol L<sup>-1</sup> s<sup>-1</sup>
Why
The coefficients of NO2 and O2 are 4 and 1. The chapter says the rate of NO2 formation is four times the rate of O2 formation. So the rate of O2 formation is 4.8 × 10-5 ÷ 4 = 1.2 × 10-5 mol L-1 s-1.
Question 8
A student in a laboratory in Addis Ababa puts marble chips and dilute HCl in a flask on a balance. The mass of the flask and its contents goes down during the reaction. Why does the mass decrease?
Answer: C Carbon dioxide gas escapes from the flask.
Why
In this experiment the reaction produces carbon dioxide gas. The gas leaves the flask, so the total mass in the flask goes down. The decrease in mass is used to measure the rate of reaction.
Question 9
For the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), at a certain moment O2 is reacting at 0.050 mol L-1 s-1. At what rate is H2O being formed?
Answer: C 0.060 mol L<sup>-1</sup> s<sup>-1</sup>
Why
Rate = -(1/5)Δ[O2]/Δt = (1/6)Δ[H2O]/Δt. So Δ[H2O]/Δt = (6/5) × 0.050 = 0.060 mol L-1 s-1. Water has the bigger coefficient, so it forms faster than O2 is used up.
Question 10
Which rate expression is correct for the reaction 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)?
Answer: A Rate = -(1/5)Δ[Br<sup>-</sup>]/Δt = -(1/6)Δ[H<sup>+</sup>]/Δt
Why
Each reactant term is divided by its own coefficient. Br- has a coefficient of 5 and H+ has a coefficient of 6. A minus sign is used for reactants because their concentrations decrease.
The app holds 12 questions on this unit, marked as you go. These are the first 10.
Unit 4 runs from page 193 to page 221 of the Ministry of Education edition.
Ministry of Education edition, English, 342 pages
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