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Chemical Reactions - Reaction Rates

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Rate of Reaction is defined as the change in concentration, mass, or volume of a reactant or product per unit time. It can be expressed as Δ[Product]Δt\frac{\Delta [\text{Product}]}{\Delta t} or −Δ[Reactant]Δt-\frac{\Delta [\text{Reactant}]}{\Delta t}.

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Collision Theory states that for a chemical reaction to occur, reactant particles must collide with each other. These collisions must have a minimum amount of energy, called the Activation Energy (EaE_a), and the particles must be in the correct orientation.

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Temperature: Increasing the temperature increases the average kinetic energy of the particles. This leads to more frequent collisions and, more importantly, a higher proportion of collisions with energy ≥Ea\ge E_a.

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Concentration and Pressure: Increasing the concentration of a solution or the pressure of a gas increases the number of particles per unit volume. This increases the frequency of successful collisions.

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Surface Area: For reactions involving a solid, increasing the surface area (e.g., using a powder instead of a single large crystal) exposes more particles to the other reactant, increasing collision frequency.

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Catalysts: A catalyst is a substance that increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy (EaE_a). The catalyst itself is not chemically changed or consumed during the reaction.

📐Formulae

Average Rate=Quantity of Product FormedTime Taken\text{Average Rate} = \frac{\text{Quantity of Product Formed}}{\text{Time Taken}}

Average Rate=Quantity of Reactant UsedTime Taken\text{Average Rate} = \frac{\text{Quantity of Reactant Used}}{\text{Time Taken}}

Rate∝Collision Frequency×e−EaRT\text{Rate} \propto \text{Collision Frequency} \times e^{-\frac{E_a}{RT}}

💡Examples

Problem 1:

In an experiment, 40 cm340\text{ cm}^3 of oxygen gas (O2O_2) was collected over a period of 20 seconds20\text{ seconds} during the decomposition of hydrogen peroxide (H2O2H_2O_2). Calculate the average rate of reaction in cm3 s−1\text{cm}^3\text{ s}^{-1}.

Solution:

Rate=40 cm320 s=2 cm3 s−1\text{Rate} = \frac{40\text{ cm}^3}{20\text{ s}} = 2\text{ cm}^3\text{ s}^{-1}

Explanation:

To find the average rate, we divide the total volume of the product formed by the total time taken for the collection.

Problem 2:

Explain why the reaction between magnesium ribbon and hydrochloric acid (HClHCl) is faster when the acid concentration is increased from 0.5 mol dm−30.5\text{ mol dm}^{-3} to 2.0 mol dm−32.0\text{ mol dm}^{-3}.

Solution:

Increasing the concentration of HClHCl increases the number of H+H^+ ions in a given volume. This leads to a higher frequency of successful collisions between the H+H^+ ions and the magnesium (MgMg) surface, thereby increasing the reaction rate.

Explanation:

Collision theory dictates that rate is proportional to the frequency of effective collisions. Higher concentration means particles are more 'crowded', making collisions more likely.

Problem 3:

A student reacts 1.0 g1.0\text{ g} of calcium carbonate (CaCO3CaCO_3) 'marbles' with excess HClHCl. In a second trial, 1.0 g1.0\text{ g} of CaCO3CaCO_3 powder is used. Which trial will reach completion first and why?

Solution:

The trial using 1.0 g1.0\text{ g} of CaCO3CaCO_3 powder will reach completion first.

Explanation:

Powder has a much larger total surface area than large marble chips. A larger surface area allows more CaCO3CaCO_3 particles to be in contact with the HClHCl solution simultaneously, increasing the collision frequency and the overall rate of reaction.