xSkill: notes, lectures and practice for CBSE and CSE

1.1 Measuring the rate A reaction rate tracks how quickly the concentration of a reactant or a product changes as time passes. For a reactant the concentration falls, so a negative sign makes the rate positive. Average rate uses a finite interval, while instantaneous rate uses the slope of the tangent at one instant. The rate law expresses the rate as a product of a rate constant and concentrations raised to powers that are found by measurement; a balanced equation does not provide them. The overall order adds up all the exponents in the rate law, and the exponent on one reactant is the order in that reactant. Molecularity belongs to an elementary step and is the number of species colliding in that step. Order can be zero, fractional or negative; molecularity is always a positive whole number.

average rate = - \frac{Δ [ R ]}{Δ t} = \frac{Δ [ P ]}{Δ t}

Rate of disappearance of R and appearance of P

rate = k [A]^{x} [B]^{y}, order = x + y

Rate law and overall order

zero order : [R] = [R]_{0} - k t

Integrated rate equation, k in mol L^-1 s^-1

zero order : t_{1/2} = \frac{[ R ] _{0}}{2 k}

Half-life of a zero order reaction, proportional to initial concentration

first order : k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}

Integrated form, k in s^-1

first order : t_{1/2} = \frac{0.693}{k}

Half-life independent of initial concentration

second order : \frac{1}{[ R ]} = \frac{1}{[ R ] _{0}} + k t

k in L mol^-1 s^-1

1.2 Temperature and activation energy The rate constant rises steeply with temperature because a larger fraction of collisions has enough energy to cross the activation barrier. The Arrhenius equation ties k, temperature and activation energy together, and its logarithmic form is a straight line when log k is plotted against 1/T. The slope gives the activation energy and the intercept gives the pre-exponential factor.

k = A e^{- E_{a} / R T}

Arrhenius equation

lo g k = lo g A - \frac{E _{a}}{2.303 R T}

Straight line form with slope -Ea/2.303R

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})

Two temperature form, R = 8.314 J K^-1 mol^-1

1.3 Units and checks The unit of k follows from the order: mol L^-1 s^-1 for zero order, s^-1 for first order and L mol^-1 s^-1 for second order. A first order half-life is a constant of the reaction, so a graph of concentration against time is exponential and a graph of log concentration against time is a straight line. When a question gives the fraction decomposed, the fraction left is one minus that fraction, and the logarithm uses amounts rather than concentrations because the ratio cancels the volume.

Exam

Identify the order from the units of k when the units are given, and from a linearity test when data are given. A plot that is linear in log[R] means first order; one linear in 1/[R] means second order.

1.4 Reading a kinetics question Most kinetics questions start with data rather than with a formula. Read first whether the question gives concentration against time, a rate table or two rate constants at two temperatures. Concentration against time needs the integrated equation, a rate table needs the initial rate method, and two rate constants need the two temperature Arrhenius form. Deciding this before writing anything saves time and prevents using the wrong equation. Always convert minutes or hours to seconds only if the answer asks for seconds; otherwise keep the time unit that the data use.

A useful habit is to write the unit of k that matches the order before substituting numbers, because a mismatch between the unit and the order exposes an error before the arithmetic is finished.