haya.Learning made clear
Unit III · Chemical kinetics
8 min readConcept sheet1-mark MCQ and a 2-mark question on writing rate expressions or finding a rate

Rate of a reaction: average and instantaneous rate

The rate of a reaction is the change in concentration per unit time. Average rate covers an interval; instantaneous rate is the slope of the tangent at one moment.

Rate of reactionRate = −Δ[R]/Δt = Δ[P]/Δt, where [R] and [P] are the molar concentrations of a reactant and a product
Unitmol L⁻¹ s⁻¹ (for gases, atm s⁻¹ when partial pressures are used)
Instantaneous rater = −d[R]/dt = d[P]/dt: the slope of the tangent to the concentration-time curve at time t
With coefficientsFor aA + bB → cC + dD: rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = (1/c) d[C]/dt = (1/d) d[D]/dt

What the rate of a reaction means

Some reactions are very fast. Silver chloride forms the moment silver nitrate meets sodium chloride solution. Some are very slow, like the rusting of iron. Others, like the inversion of cane sugar, go at a moderate speed. Chemical kinetics is the branch of chemistry that studies these speeds and how reactions happen. The rate of a reaction is the change in concentration of a reactant or a product in unit time. Square brackets mean molar concentration: [R] is the concentration of R in mol L⁻¹. A reactant is used up, so Δ[R] is negative. We put a minus sign before it so that the rate is always positive.

Average rate vs instantaneous rate

Average rateThe rate over a time interval: r(av) = −Δ[R]/Δt = −([R]₂ − [R]₁)/(t₂ − t₁). On a graph of [R] against t, it is the slope of the straight line (chord) joining the two points. It changes with the interval you choose.
Instantaneous rateThe rate at one moment, found when Δt becomes very small: r(inst) = −d[R]/dt. On the graph, draw a tangent to the curve at time t; the rate is minus its slope (for a product curve, it is the slope itself).

The rate usually falls as the reaction goes on, because the reactant concentration falls. So the curve of [R] against t gets flatter with time.

How to draw the concentration-time graph

  • Draw two axes. Label the x-axis Time (s) and the y-axis [R] (mol L⁻¹).
  • Draw a curve that starts high on the y-axis at [R]₀ and falls, getting flatter as time goes on.
  • For the average rate, mark two points at t₁ and t₂ and join them with a straight line. Label Δ[R] (vertical side) and Δt (horizontal side).
  • For the instantaneous rate, draw a tangent that touches the curve at one time t. Label it "slope = d[R]/dt".
  • You may draw the product curve [P] on the same axes: it starts at zero and rises, getting flatter.

Rate when the coefficients are not 1

In 2HI(g) → H₂(g) + I₂(g), two moles of HI are used for every mole of H₂ formed. So HI disappears twice as fast as H₂ appears. To get one rate for the whole reaction, divide each rate term by its coefficient: Rate = −½ d[HI]/dt = d[H₂]/dt = d[I₂]/dt. In the same way, for 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O: Rate = −(1/5) d[Br⁻]/dt = −d[BrO₃⁻]/dt = −(1/6) d[H⁺]/dt = (1/3) d[Br₂]/dt = (1/3) d[H₂O]/dt.

Worked example: average rate of 2HI → H₂ + I₂

Problem

In the reaction 2HI(g) → H₂(g) + I₂(g), the concentration of HI falls from 0.80 mol L⁻¹ to 0.62 mol L⁻¹ in 15 minutes. Find (a) the average rate of reaction and (b) the rate of formation of H₂, in mol L⁻¹ min⁻¹ and in mol L⁻¹ s⁻¹.

Working
  1. Given: [HI]₁ = 0.80 mol L⁻¹, [HI]₂ = 0.62 mol L⁻¹, Δt = 15 min.
  2. Δ[HI] = 0.62 − 0.80 = −0.18 mol L⁻¹.
  3. Formula: Rate = −½ Δ[HI]/Δt.
  4. Rate = −½ × (−0.18 mol L⁻¹) ÷ 15 min = 0.09 ÷ 15 = 6.0 × 10⁻³ mol L⁻¹ min⁻¹.
  5. Rate of formation of H₂ = Δ[H₂]/Δt = rate of reaction = 6.0 × 10⁻³ mol L⁻¹ min⁻¹.
  6. In seconds: 6.0 × 10⁻³ ÷ 60 = 1.0 × 10⁻⁴ mol L⁻¹ s⁻¹.
Answer

(a) and (b) both 6.0 × 10⁻³ mol L⁻¹ min⁻¹ = 1.0 × 10⁻⁴ mol L⁻¹ s⁻¹

Worked example: linking the rates of different species

Problem

For 2N₂O₅(g) → 4NO₂(g) + O₂(g), N₂O₅ disappears at 1.6 × 10⁻⁴ mol L⁻¹ s⁻¹. Find the rate of reaction and the rates of formation of NO₂ and O₂.

Working
  1. Rate = −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt.
  2. Rate of reaction = ½ × 1.6 × 10⁻⁴ = 8.0 × 10⁻⁵ mol L⁻¹ s⁻¹.
  3. d[NO₂]/dt = 4 × 8.0 × 10⁻⁵ = 3.2 × 10⁻⁴ mol L⁻¹ s⁻¹.
  4. d[O₂]/dt = 1 × 8.0 × 10⁻⁵ = 8.0 × 10⁻⁵ mol L⁻¹ s⁻¹.
Answer

Rate of reaction 8.0 × 10⁻⁵; NO₂ forms at 3.2 × 10⁻⁴; O₂ forms at 8.0 × 10⁻⁵ (all in mol L⁻¹ s⁻¹)

In real life

Milk turns to curd faster on a hot day in Chennai than in a cold Shimla winter, and a refrigerator keeps food fresh longer. Both are about controlling the rate of chemical reactions.

Exam angle

2 marks: “Define average rate and instantaneous rate of a reaction. How is the instantaneous rate found from a graph?” Average rate = change in concentration ÷ time interval, −Δ[R]/Δt. Instantaneous rate = rate at one moment, −d[R]/dt. Draw a tangent to the [R]-t curve at that time; the rate is minus its slope.

2 marks: “For N₂ + 3H₂ → 2NH₃, NH₃ forms at 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹. Find the rates of disappearance of N₂ and H₂.” Rate = ½ d[NH₃]/dt = 1.0 × 10⁻⁴. So −d[N₂]/dt = 1.0 × 10⁻⁴ mol L⁻¹ s⁻¹ and −d[H₂]/dt = 3 × 1.0 × 10⁻⁴ = 3.0 × 10⁻⁴ mol L⁻¹ s⁻¹.

1 marks: “Why is a minus sign put before Δ[R] in the rate expression?” [R] decreases, so Δ[R] is negative; the minus sign makes the rate a positive quantity.

Remember it

Reactants get a minus, products a plus, and every term is divided by its own coefficient.