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Grade 11-12 (age 16-18)

high school Reaction Rate

The same reaction is slow in a refrigerator and fast in the sun; reaction rate is how quickly reactants are used up or products formed per unit time. Molecules carry different kinetic energies, and only those that collide with more than the activation energy Eₐ can react. Raising the temperature greatly increases the share of high-energy molecules, and a catalyst opens a new route with a lower Eₐ so that more molecules react. The effect of concentration is written as the rate law v = k[A]^m[B]^n, where the orders m and n are found by experiment. Changing the temperature, Eₐ and the catalyst in the diagrams changes the share of molecules that can react, shown as a number.

Why Are Some Reactions Slow?

💡 Climbers Crossing a Pass
①For a reaction to happen, molecules must collide hard enough to cross a pass called the activation energy Eₐ
②Even at one temperature, molecules have a spread of kinetic energies, so only some of them make it over
③The higher the temperature, the more fast molecules there are and the more cross the pass
④That is why food spoils more slowly in a refrigerator
⑤Reaction rate is the amount of reactant used up, or product formed, per unit time

Temperature and the Energy Distribution

25 °C
50
🔍 What the Diagram Shows
①The area under each curve is the total number of molecules, so it stays the same as the temperature changes
②Raising the temperature makes the curve lower and wider, with a thicker tail on the right
③The shaded area to the right of the Eₐ line is the molecules that can react, and its share is the number on the bottom line
④Raising Eₐ shrinks the shaded area quickly
⑤To show the shape, the diagram sets Eₐ lower than in real reactions. In a real reaction Eₐ lies much farther out in the tail, so a 10 °C rise near room temperature roughly doubles the share of molecules beyond it

The Effect of a Catalyst

without catalyst
💡 What a Catalyst Does
①A catalyst opens a new reaction route with a lower Eₐ
②With a lower pass, many more molecules can cross it at the same temperature, as the two shares on the bottom line of the diagram show
③The catalyst is still there after the reaction; it is not used up
④It lowers Eₐ for the forward and reverse reactions alike, so ΔH and the position of equilibrium do not change

The Rate Law

Rate Law
v = k[A]m[B]n
k: rate constant, m and n: reaction orders (found by experiment)
Arrhenius Equation
k = Ae-E_a/RT
A higher temperature or a lower Eₐ makes k larger and the reaction faster
📐 Concentration and Rate
①A higher reactant concentration means more collisions and a faster reaction
②The orders m and n are found only by experiment, regardless of the coefficients in the equation
③A zero-order reaction does not depend on concentration, a first-order reaction is proportional to it, and a second-order reaction to its square
④The half-life (t½) is the time it takes for a reactant to fall to half of its starting amount

Worked Examples

Example 1
For a reaction with rate law v = k[A][B], how does the rate change if only [A] is doubled?
1
The order with respect to [A] is 1, so the rate is directly proportional to [A].
v = k[A][B], first order in [A]
2
Doubling [A] doubles the rate.
v ∝ [A] ⇒ × 2
▸ It doubles
The exponents in the rate law are the orders; with an order of 1, doubling the concentration doubles the rate.
Example 2
In a reaction, doubling [A] makes the rate four times faster. What is the order with respect to A?
1
The rate is proportional to [A]m, so doubling [A] multiplies the rate by 2m.
2m = 4
2
Find the m that satisfies 2m = 4.
2m = 4 = 22 ⇒ m = 2
▸ Second order (m = 2)
Reaction orders come from experimental results like this, not from the coefficients in the equation: see how many times faster the rate became and match it to 2m.

Summary

Key Formula
v = k[A]m[B]n
The rate constant k depends on temperature and activation energy
CSAT-style problem, adapted
Which statement about a catalyst is INCORRECT?
①It lowers the activation energy to speed up the reaction
②It speeds up both the forward and reverse reactions
③It is not consumed during the reaction
④It changes the equilibrium constant K and the equilibrium position
⑤It provides a new reaction pathway
▸ ④ It changes the equilibrium constant K and the equilibrium position
1
A catalyst lowers the activation energy of the forward and reverse reactions together; it only helps the system reach equilibrium sooner.
2
K and the position of equilibrium do not change, so ④ is not correct.
🎯 Exam Points
①In v = k[A]^m[B]^n, m and n are found only by experiment and have nothing to do with the coefficients
②A higher temperature raises the share of molecules above Eₐ, so the reaction speeds up
③A catalyst speeds a reaction by lowering Eₐ but changes neither ΔH nor the position of equilibrium
④Higher concentration means more frequent collisions and a faster reaction
⑤The half-life of a first-order reaction is constant, whatever the starting concentration (the same pattern as radioactive decay)
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Electrochemistry

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