Reaction Rate

Reaction rate is how fast reactants fall or products rise per unit time, and only molecules with energy above the activation barrier can react. Higher temperature spreads the energy distribution so more molecules react, and a catalyst speeds the reaction by lowering Ea. In the rate law v = k[A]^m[B]^n the orders come from experiment, and concentration, temperature, and catalysts control the rate. Here you change temperature, Ea, and catalyst on/off to compare the distribution curve and the energy path.

Why Are Some Reactions Slow?

💡 Analogy: Climbing a Mountain

①Every reaction must climb an 'energy hill' (E_a)

②Molecules have kinetic energy, but not all the same speed

③Only molecules with energy above E_a can react

④Refrigerating food = lower T → less molecular motion → slower decay

⑤Reaction rate = decrease of reactants per unit time (or increase of products)

Temperature and Molecular Energy Distribution

Temperature (°C)25

Activation Energy E_a (kJ)50

🔍 Watch the Shaded Area!

①Higher T → distribution shifts toward higher energy

②Fraction with E ≥ E_a grows substantially

③Raising T by 10°C ≈ 2~4× faster

④Lower E_a → temperature effect is relatively smaller

Effect of a Catalyst

Catalyst (0=no, 1=yes)0

💡 How Catalysts Work

①Catalyst provides a new path → lowers E_a

②More molecules can cross E_a → faster reaction

③Catalyst is unchanged before and after (not consumed)

④Lowers both forward and reverse E_a → equilibrium position unchanged

Rate Law

v = k[A]^m[B]ⁿ

k: rate constant; m, n: reaction orders (determined by experiment)

Arrhenius Equation

k = Ae^-E_a/RT

T↑ → k↑ → faster | E_a↓ → k↑ → faster

📐 Concentration and Rate

①[A]↑ → more collisions → rate↑

②Orders m, n are determined experimentally (not from coefficients!)

③Zero-order: rate independent of concentration; 1st-order: proportional; 2nd-order: ∝ [A]²

④Half-life (t½): time for reactant to drop to half

Worked Examples

Example 1

For a reaction with rate law v = k[A][B], by what factor does the rate change if [A] is doubled? (other conditions constant)

The order in [A] is 1, so the rate is directly proportional to [A].

v = k[A][B], first order in [A]

Doubling only [A] → rate doubles.

v ∝ [A] ⇒ 2×

▸ 2×

The exponents in the rate law are the orders. Order 1 in [A] → doubling [A] doubles the rate.

Example 2

Doubling [A] makes the rate 4×. What is the reaction order with respect to A?

From rate ∝ [A]^m, doubling [A] multiplies the rate by 2^m.

2^m = 4

Solve for m.

2^m = 4 = 2² ⇒ m = 2

▸ Second order (m = 2)

Reaction order is determined by experiment, not by stoichiometric coefficients. Rate factor = 2^m.

Summary

Key Formula

v = k[A]^m[B]ⁿ

Rate constant k depends on T and E_a

CSAT-style

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

A catalyst lowers the activation energy of forward and reverse equally, only reaching equilibrium faster.

K and the equilibrium position do not change, so ④ is incorrect.

🎯 Exam Points

①v = k[A]^m[B]^n — m, n only by experiment

②T↑10°C → ~2–4× faster (Arrhenius)

③Catalyst lowers E_a (equilibrium unaffected)

④Higher concentration → more collisions → faster

⑤1st-order half-life: t½ independent of concentration

← Previous

Equilibrium Shift

Electrochemistry

Was this helpful? Support seegongsik