Interpret data, including graphs, from rate of reaction experiments

Chemical Reactions – Rate of Reaction

What is Reaction Rate? 🔬

Think of a reaction like a busy highway. The reaction rate is how many cars (molecules) pass a point per unit time. A fast reaction is like a rush‑hour traffic jam – lots of cars moving quickly. A slow reaction is like a quiet road – only a few cars moving slowly.

Factors That Change the Rate ⚡

  • Concentration – More reactants = more collisions.

  • Temperature – Hotter molecules move faster, collide more often.

  • Surface Area – Grinding a solid into powder gives more surface for collisions.

  • Catalyst – Provides an easier path, like a shortcut on the highway.

  • Pressure – For gases, higher pressure pushes molecules together.

Rate Law – The Mathematical Formula \$rate = k[A]^m[B]^n\$

k is the rate constant (depends on temperature). m and n are the reaction orders for reactants A and B. They tell us how the rate changes when we change concentrations.

Example: For the synthesis of ammonia:

\$\ce{N2 + 3H2 -> 2NH3}\$

The rate law is often written as:

\$rate = k[\ce{N2}]^1[\ce{H2}]^3\$

(orders 1 and 3, respectively).

Measuring Reaction Rate 📈

  1. Concentration Method – Measure concentration of a reactant or product at different times.

  2. Volume Method – For gas reactions, measure volume change (e.g., gas evolution).

  3. Time Method – Record the time taken for a fixed change in concentration.

Tip: Always use the same units (e.g., M for molarity, s for seconds) to avoid confusion.

Interpreting Graphs 📊

Concentration vs. Time

A steep slope means a fast reaction. A shallow slope means a slow reaction. The slope at any point is the instantaneous rate.

Rate vs. Concentration

Plotting rate against concentration of one reactant (keeping others constant) gives a straight line if the reaction is first order in that reactant. The slope is the rate constant times the concentration of the other reactants raised to their orders.

Rate vs. Temperature

An Arrhenius plot (ln k vs. 1/T) is a straight line. The slope equals –Eₐ/R (activation energy over gas constant). A steeper slope indicates a higher activation energy.

Sample Data Table

Time (s)[A] (M)Rate (M s⁻¹)
00.100-
300.080-6.7×10⁻⁴
600.065-5.0×10⁻⁴

Interpretation: The rate decreases as concentration falls, indicating a first‑order dependence on A.

Example Problem 🚀

A reaction follows the rate law \$rate = k[\ce{C}]^2\$. In an experiment, the rate is \$1.2\times10^{-3}\,\text{M s}^{-1}\$ when \$[\ce{C}] = 0.05\,\text{M}\$. What is the rate constant \$k\$?

  1. Write the rate law with the given values: \$1.2\times10^{-3} = k(0.05)^2\$.

  2. Calculate \$(0.05)^2 = 2.5\times10^{-3}\$.

  3. Solve for \$k\$: \$k = \dfrac{1.2\times10^{-3}}{2.5\times10^{-3}} = 0.48\,\text{M}^{-1}\text{s}^{-1}\$.

Result: \$k = 0.48\,\text{M}^{-1}\text{s}^{-1}\$.

Exam Tips for IGCSE Chemistry 0620

  • Always state the units of your rate constant.

  • When interpreting graphs, identify the slope and relate it to the rate.

  • Remember that the rate law is derived from experimental data, not theory.

  • Use the Arrhenius equation to calculate activation energy if required.

  • Check that your answer includes the correct significant figures.

Good luck! 🎓 Remember: practice makes perfect – keep plotting graphs and solving rate law problems.