Unit 7: Reaction Kinetics — Short Questions

Ans. Rate of reaction is the change in concentration of a reactant or product per unit time. It shows how fast or slow a chemical reaction occurs.

Ans.
Property: Enthalpy Change (ΔH)
Definition: Heat absorbed or released during a reaction
Depends on: Initial and final energy states
Symbol: ΔH

Property Activation Energy (Ea)
Definition: Minimum energy required to start the reaction
Depends on: Energy barrier between reactants and products
Symbol: Ea

Ans.
Order of Reaction: Determined experimentally, Can be zero, fractional, or whole number, Applies to overall reaction
Molecularity: Based on reaction mechanism, Always a whole number (1, 2, or 3), Applies to elementary step only

Ans. Because the concentration of reactants decreases over time, reducing the frequency of collisions and slowing down the reaction rate.

Ans.

• Temperature: Increases kinetic energy gives more frequent and energetic collisions and faster rate.
• Surface area: Greater surface area (e.g., powdered solid) gives more area for collisions and increased reaction rate.

Ans. In radioactive decay, the rate depends only on the amount of radioactive substance present, not on its concentration with other substances. Hence, it follows first-order kinetics:

Ans. If a reaction is second order:

• Doubling concentration gives rate increases by 4 times (2²)
• Concentration reduced to half gives rate decreases to 1/4th

Ans. Half-life (t1/2) is the time required for half the reactant to decay or react.
Used to determine:

• Speed of radioactive decay
• Stability of substances
• Kinetics of first-order reactions

Ans. Because pure oxygen increases the concentration of O2 than air, leading to more frequent and effective collisions and faster combustion.

Ans. In an energy profile diagram, a catalyst provides an alternative pathway with lower activation energy, making it easier for reactants to convert into products.

Ans. Rate constant and order
Given: k = 3.5 × 10⁻⁴ s⁻¹
The unit "s⁻¹" implies it is a first-order reaction (as per standard units for rate constants).

Ans. Initial concentration = 0.50 mol/dm³,
k = 3.5×10⁻⁴ s⁻¹
For first-order reaction:
Rate = k [A] = (3.5×10⁻⁴)(0.50) = 1.75×10⁻⁴ mol dm⁻ s⁻¹

Ans. Effect of doubling concentration on rate
Since it's first order, rate is directly proportional to concentration:
Rate doubles when concentration is doubled.
Rate = k[A]

Ans. Half-life calculation
Given: k = 2.5×10⁻³ s⁻¹
t1/2 = 0.693 / k = 0.693 / (2.5×10⁻³) = 2772.2 seconds
Rate constant from half-life
Given: t1/2 = 12 hrs = 43200 s
t1/2 in minutes = 2772.2/60 = 4.62 minutes

Ans. To calculate the rate constant (k) for a first-order radioactive decay process, use the formula:
k = 0.693 / t1/2
Given: Half-life t1/2 = 12 hours
Convert hours to seconds:
12 hours = 12 × 60 × 60 = 43,200 seconds
Plug into the formula:
k = 0.693 / 43200 = 1.60×10⁻⁵ s⁻¹
Answer
k = 160×10⁻⁵ s⁻¹

Ans. Reaction kinetics is the study of the rates of chemical reactions and the factors affecting those rates. It also explains the mechanism by which a reaction occurs.
Example: Decomposition of hydrogen peroxide (H2O2) into water and oxygen is studied to find its rate and mechanism.

Ans. It is the change in concentration of reactants or products per unit time.

Ans. Rate=-d[A]/dt=-d[B]/dt=d[C]/dt

Ans. The units are usually mol dm⁻³ s⁻¹, representing moles formed or used per liter per second.

Ans.

• Average rate is calculated over a time interval Avg. Rate=Δ[Concentration]/Δt
• Instantaneous rate is the rate at a particular moment (slope of concentration vs. time curve). Example: Graph of H2O2 decomposition helps estimate instantaneous rate at 20 s.

Ans. Increasing temperature increases the kinetic energy of molecules, causing more effective collisions, thus increasing the rate.
Example: Rate of rusting increases in summer due to higher temperature.

Ans. Higher concentration means more particles in a given volume, increasing collision frequency, thus increasing the rate.
Example: Strong HCl reacts faster with Zn than dilute HCl.

Ans. A larger surface area (finely divided solids) provides more area for collisions, increasing the rate.
Example: Powdered magnesium burns faster than ribbon.

Ans. A catalyst increases the rate by providing an alternative pathway with lower activation energy, without being consumed.
Example: MnO2 speeds up H2O2 decomposition.

Ans. It states that for a reaction to occur, particles must collide with proper orientation and sufficient energy (activation energy).
Example: CH4 + Cl2 reacts in sunlight due to energetic collisions.

Ans. The minimum energy that reacting particles must have to form the activated complex and convert into products.
Example: In combustion of wood, heat provides activation energy to start burning.

Ans. It provides a new reaction path that requires less energy for the same reaction to proceed.

Ans. It is a step-by-step description of a chemical reaction showing how reactants convert into products through elementary steps.
Example
For: 2NO2 → 2NO + O2
Mechanism:
1. NO2 → NO + O (slow)
2. NO2 + O → NO + O2 (fast)

Ans. The slowest step in a reaction mechanism that determines the overall rate.
Example: In the above NO2 mechanism, step 1 is rate-determining.

Ans. It is the number of reactant molecules involved in an elementary step.
Types: Unimolecular (1), Bimolecular (2), Termolecular (3)
Example
H2 → 2H (unimolecular)
H2 + Cl2 → 2HCl (bimolecular)

Ans. It is the sum of powers of concentration terms in the rate law.
Example
Rate = k[A][B]² → order = 1 + 2 = 3

Ans. By observing how changing concentrations affect the rate and comparing using rate laws.

Ans. Rate=k[A]ᵐ[B]ⁿ

• k = rate constant
• m, n = orders with respect to A and B
• A, B = concentrations of reactants

Ans. Units of rate constant are:

• Zero order: mol dm⁻³ s⁻¹
• First order: s⁻¹
• Second order: mol⁻¹ dm³ s⁻¹

Ans.
1. Volume of gas evolved
2. Color change or turbidity
Example
Measuring O2 evolved during H2O2 decomposition.

Ans. By finding the slope (tangent) of the curve at a given time.

Ans.

• Homogeneous catalyst: Same phase of reactants, product and catalyst Example: H in ester hydrolysis
• Heterogeneous catalyst: Different phase Example: Ni solid in hydrogenation of alkenes

Ans. Amylase catalyzes breakdown of starch to maltose in the mouth (biological catalysis).

Ans. They increase production rate, reduce energy costs, and improve yield.
Example: Iron catalyst in Haber process.

Ans. It shows energy changes during a reaction, with activation energy and enthalpy difference.

Ans. It lowers the peak (activation energy) without changing reactants/products' energy.

Ans. Yes. Kinetics (rate) and thermodynamics (feasibility) are different.
Example: Diamond to graphite is thermodynamically favored but very slow.

Ans. Used in drug design to control the rate of drug reactions or breakdown in the body.

Ans. Activation energy (Ea) is the minimum energy required for reactant molecules to form the activated complex and proceed to products.
It is like an energy barrier that must be crossed for a reaction to occur.
Role: Activation energy controls whether a reaction will happen at a given temperature. Molecules must have at least this much energy to react.

Ans. See Q.3 of theory.

Ans. See Q.10 of theory.

Ans. See Q.5 of theory.

Ans. (i) The overall balance equation for the reaction is given below:
H2O2 + 2I⁻ + 2H⁺ → 2H2O + I2
(ii) The intermediate for the reaction is HOI and there is no catalyst in reaction mechanism.
(iii) The step 1 is called the rate determining step as it is a slow step.
(iv) The rate equation is derived from the slow step and it is given below:
Rate = k [H2O2] [I⁻]

Solution We are given:

• Rate law: Rate=k[A][B]² [A] = 0.5 mol/L [B] = 0.2 mol/L k = 4.0mol⁻²s⁻¹

Calculation
Rate = 4.0 × (0.5) × (0.2)²
Rate = 4.0 × 0.5 × 0.04 = 0.08 mol. L⁻¹.S⁻¹

Ans.
Rate = 0.08mol L⁻¹ s⁻¹

Temp. (K) | Rate constant (cm³ mol⁻¹ s⁻¹) (k)
500 | 6.814 x 10⁻⁴

Solution To find the half-life (t1/2) of a first-order reaction, we use the formula:
t1/2 = 0.693 / k

Given
k=5.5×10⁻¹⁴ s⁻¹

Calculation
t1/2 = 0.693 / (5.5×10⁻¹⁴) = 1.26×10¹³ seconds

To convert into years t1/2 = 1.26×10¹³ / (60×60×24×365) ≈ 399,365 years

Ans.
t1/2 = 1.26×10¹³ seconds ≈ 3.99×10² years

Experiment | [HI] (M) | Rate (M/s)
1 | 0.015 | 1.1 × 10⁻³
2 | 0.030 | 4.4 × 10⁻³
3 | 0.045 | 9.9 × 10⁻³

Write the rate law for the reaction. Find the value and units of the specific rate constant, k.

Solution

Given Reaction 2HI(g) → H2(g) + I2(g)

Step 1 Determine the Order of the Reaction

Compare Experiments 1 and 2 [HI] doubles: 0.015 → 0.030
Rate increases: 1.1 × 10⁻³ → 4.4 → 10⁻³

Rate₂/Rate₁ = (4.4×10⁻³) / (1.1×10⁻³) = 4

([HI]₂/[HI]₁)ⁿ = (0.030/0.015)ⁿ = 2ⁿ

So, the reaction is second order with respect to HI.

Step 2 Write the rate law
Rate=k[HI]²

Step 3 Find the rate constant k

Use Experiment 1 Rate = 1.1 × 10⁻³ M/s, [HI] = 0.015 M
1.1 × 10⁻³ = k (0.015)² ⇒ k = 1.1×10⁻³ / (2.25×10⁻⁴) = 4.89 M⁻¹s⁻¹

Answers
Rate law
Rate = k [HI]²

Rate constant k = 4.89 M⁻¹s⁻¹