STPM Chemistry Form 6 Notes – Reaction Kinetics (Part 3)

by BerryBerryTeacher

in Berry Reference (Notes)

In this Part 3 of STPM Form 6 Chapter on Reaction Kinetics from Berry Berry Easy, you’ll be learning the basics on collision theory, rate constant and rate law,first order reactions, second order reactions and zero order reactions. Please pay utmost attention to what is taught in this post as the knowledge presented here are going to be tested in one form or another in your exams. However, before you continue on to the post, please ensure that you have already mastered the definition in Part 1 and understand the underlying concepts in Part 2.

(Tips: Try out all 4 examples in this post to ensure that you fully understand what is shown in this post. Students might also be under false impression that they already understand everything about the collision theory and rate constants which might have been expose to Berry Readers during your SPM period. Try to learn it anew as if you have not seen it before.)

STPM Chemistry Form 6 Notes – Reaction Kinetics (Part 3)

Collision Theory

  • Reaction – reactants particle must collide with one another with sufficient energy to break chemical bonds in the reactants to form product.
  • Activated complex – a very high energetic and highly unstable species is formed.
  • Chemical reaction – is the effective collisions of reactant particles.
  • Reaction rate – is the measurement of the frequency of effective collisions.
  • Activation energy (Ea) – minimum energy required to break the chemical bonds in the reactant molecules and overcome the repulsion forces of the reactants molecules.
  • Energy profile / Reaction profile – the difference in energy between the reactants and activated complex.

The rate of a reaction is affected by

  • temperature (reaction rate increases with increasing temperature)
  • physical state of reactants (gas, liquid or solid)
  • activation energy (reaction rate increases with increasing temperature)
  • catalysts (reaction rate increases with positive catalysts and reaction rate decreases with negative catalysts)
  • solvent (solvents affect the transition state stability)
  • collision frequency
  • collision orientation
  • the concentration of the reactants in the rate determining step
  • effect of pressure

Example 1

Which of the following does NOT always affect the rate of reaction?

A. Changing the temperature
B. Adding a catalyst
C. Increasing the volume by adding solvent
D. Adding a reactant

Answer: D

Solution: Changing temperature changes the reaction rate. Adding a catalyst (usually positive catalyst) lowers the activation energy which results increases the reaction rate. Increasing the volume reduces the concentration of all reagents (include the reactants in the rate determining step). Increasing volume of a gas phase reaction results in a decreased in reaction rate. Therefore, if the reactant being added is not involved in the rate determining step, then it does not influence the rate.

Rate Constant and Rate Law

Rate constant is determined by the Arrhenius constant (collision orientation and frequency take into account)

k = A e –Ea/RT

where Ea is the activation energy, R is the molar gas constant, T is the absolute temperature and A is the frequency factor.

ln k = ln A – (Ea/RT)
lg k = lg A – (Ea/2.303R) (1/T)

A plot of lg k against 1/T will be a linear graph with a slope of – (Ea/2.303R) and an intercept of lg A

First Order Reactions

The rate equation is
rate = k [A]

Example 2

Rate expression for the decomposition of N2O5,
rate = k [N2O5] or –d[N2O5]/dt = k [N2O5]1
Unit of k = mol dm-3 s-1 / mol dm-3 = s-1

ln [N2O5]t = ln [N2O5]0 – kt , where ln = natural logarithm
For the first-order reaction, a plot of ln [N2O5] against t will be linear with a slope of –k and an intercept of ln [N2O5]0.

or

ln {[N2O5]0 / [N2O5]} = kt
For the first-order reaction, a plot of ln {[N2O5]0 / [N2O5]} against t will be straight line with a slope of k and an intercept through the origin (0).

Second Order Reactions

The rate equation is
rate = k [A]2 or rate = k [A] [B]

Example 3

2NO2(g)  –> 2NO(g) + O2(g)
rate = k [NO2]2
Unit of k = mol dm3 s-1/ (mol dm3 s-1)2 = mol dm3 s-1

1/[N2O5]t = kt + 1/[ N2O5]0
For the second-order reaction, a plot of 1/[N2O5] against t will be a slope of k and an intercept of 1/[ N2O5]0

Zero Order Reaction

The rate equation is
rate = k [A]0

Example 4

Thermal decomposition of hydrogen iodide on gold:

HI(g)  –> ½ H2(g) + ½ I2(g)
rate = k [NO2]2
Unit of k = mol dm-3 s-1

[HI] = [HI]0 – kt
For the zero-order reaction, a plot of [HI] against t will be a straight line with a slope of –k and an intercept of [HI]0

In the next and final post in this series of notes on Reaction Kinetics for STPM Form 6 Chemistry by Berry Berry Easy, you’ll learn all about catalysis (homogeneous and heterogeneous) and effect of temperature on reaction rates and rate constants (including kinetics theory, Arrhenius equation and Clausius-Clapeyron equation)

(homogeneous and heterogeneous)

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