Chemical Kinetics & Mechanisms • Step-by-step solutions
$$\text{Rate} = k[A]^m[B]^n$$
The rate law equation describes how the rate of a chemical reaction depends on the concentrations of reactants raised to specific powers. This fundamental equation in chemical kinetics defines the relationship between reaction rate and reactant concentrations through the rate constant (k) and reaction orders (m, n).
Key relationships:
This equation is essential for understanding reaction mechanisms, predicting reaction behavior, designing chemical processes, and studying enzyme kinetics. The reaction orders reveal the molecularity of the rate-determining step and provide insights into the reaction mechanism.
Rate depends on concentration raised to reaction order powers
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Enter parameters to see solution steps.
General form:
$$\text{Rate} = k[A]^m[B]^n$$
For a general reaction: aA + bB → Products
Overall order = m + n
Units of rate constant:
$$k \text{ has units of } \frac{\text{M}^{1-(m+n)}}{\text{s}}$$
Examples:
Zero order: Rate = k (units: M/s)
First order: Rate = k[A] (units: s⁻¹)
Second order: Rate = k[A]² or k[A][B] (units: M⁻¹s⁻¹)
The rate law equation describes the relationship between the rate of a chemical reaction and the concentrations of its reactants. It has the general form Rate = k[A]^m[B]^n, where k is the rate constant, [A] and [B] are reactant concentrations, and m and n are reaction orders. The reaction orders must be determined experimentally and cannot be deduced from the balanced chemical equation.
The complete mathematical formulation of the rate law equation:
Where:
Key applications of rate law equations:
Key concepts in reaction kinetics:
\(\text{Rate} = k[A]^m[B]^n\)
Zero: [A] = [A]₀ - kt; First: ln[A] = ln[A]₀ - kt; Second: 1/[A] = 1/[A]₀ + kt
Where k = rate constant, [A]₀ = initial concentration.
Zero: Rate = k; First: Rate = k[A]; Second: Rate = k[A]² or k[A][B]
A reaction has the rate law: Rate = k[A]²[B]. If [A] = 0.2 M, [B] = 0.3 M, and k = 0.5 M⁻²s⁻¹, what is the reaction rate?
Using the rate law: Rate = k[A]²[B]
Given: k = 0.5 M⁻²s⁻¹, [A] = 0.2 M, [B] = 0.3 M
Step 1: Rate = (0.5 M⁻²s⁻¹)(0.2 M)²(0.3 M)
Step 2: Rate = (0.5 M⁻²s⁻¹)(0.04 M²)(0.3 M)
Step 3: Rate = (0.5)(0.04)(0.3) M⁻²⁺²⁺¹s⁻¹
Step 4: Rate = 0.006 M/s
The answer is C) 0.006 M/s.
This problem demonstrates how to use the rate law equation. Remember to raise each concentration to its respective power (order). The units must also be consistent: rate units are always M/s, and the rate constant units are chosen to make the equation dimensionally consistent. In this case, since the overall order is 3, the rate constant has units of M⁻²s⁻¹.
Rate Law: Mathematical expression relating rate to concentrations
Reaction Order: Exponent in rate law expression
Rate Constant: Proportionality constant in rate law
• Rate = k[A]^m[B]^n
• Raise concentrations to their orders
• Units must be consistent
• Square [A] for second order in A
• Always check units consistency
• Rate always has units of M/s
• Forgetting to raise concentrations to powers
• Using wrong units for rate constant
• Confusing rate law with balanced equation
Explain how reaction orders are determined experimentally and why they cannot be deduced from the balanced chemical equation. For a reaction A + B → C, if doubling [A] doubles the rate while doubling [B] has no effect, what are the orders with respect to A and B, and what is the overall order?
Step 1: If doubling [A] doubles the rate, the order with respect to A is 1 (first order).
Step 2: If doubling [B] has no effect, the order with respect to B is 0 (zero order).
Step 3: Overall order = 1 + 0 = 1
Step 4: Rate law is Rate = k[A]¹[B]⁰ = k[A]
Reaction orders are determined by comparing rates at different concentrations. They reflect the molecularity of the rate-determining step and cannot be predicted from stoichiometry.
Reaction orders must be determined experimentally because they reflect the molecular mechanism of the reaction, not the stoichiometry. The balanced equation shows the overall transformation, but the rate law reveals the rate-determining step. For example, if doubling [A] doubles the rate, the rate is proportional to [A]¹, so the order with respect to A is 1. This tells us that A participates in the rate-determining step.
Rate-Determining Step: Slowest step in mechanism
Elementary Step: Individual molecular event
Molecularity: Number of molecules in elementary step
• Orders determined experimentally
• Not from balanced equation
• Reflect rate-determining step
• Compare rates at different concentrations
• Use ratio method for determining orders
• Zero order = no effect on rate
• Using stoichiometric coefficients as orders
• Not understanding experimental determination
• Confusing rate law with equilibrium expression
A first-order reaction has a rate constant of 0.005 s⁻¹. Calculate the half-life of the reaction. How long will it take for 75% of the reactant to be consumed? (Note: For first-order reactions, t₁/₂ = ln(2)/k)
Step 1: Calculate half-life using t₁/₂ = ln(2)/k
Step 2: t₁/₂ = 0.693/0.005 s⁻¹ = 138.6 s
Step 3: For 75% consumed, 25% remains, so [A] = 0.25[A]₀
Step 4: For first-order: ln([A]₀/[A]) = kt
Step 5: ln(1/0.25) = ln(4) = 1.386 = kt
Step 6: t = 1.386/0.005 = 277.2 s
The half-life is 138.6 s, and it takes 277.2 s for 75% consumption.
This problem demonstrates the special properties of first-order reactions. The half-life is constant and independent of initial concentration. For 75% consumption, 25% remains, which is (1/2)² of the original amount, meaning two half-lives have passed. Indeed, 277.2 s is approximately 2 × 138.6 s, confirming our calculation.
Half-life: Time for concentration to halve
First-Order Reaction: Rate depends linearly on one concentration
Integrated Rate Law: Relates concentration to time
• t₁/₂ = ln(2)/k for first order
• Half-life constant for first order
• ln([A]₀/[A]) = kt
• Remember: t₁/₂ = 0.693/k for first order
• For nth half-life: [A] = [A]₀/2ⁿ
• ln(2) ≈ 0.693
• Using wrong half-life formula for order
• Forgetting ln(2) in denominator
• Not accounting for logarithm properly
Consider the reaction: NO₂ + CO → NO + CO₂. The experimentally determined rate law is Rate = k[NO₂]². Based on this rate law, what can you conclude about the reaction mechanism? Could this reaction occur in a single elementary step?
Step 1: The rate law shows second order dependence on [NO₂] and zero order on [CO].
Step 2: This means [CO] does not appear in the rate-determining step.
Step 3: The reaction cannot occur in a single elementary step because that would require both NO₂ and CO in the rate law.
Step 4: The mechanism likely involves multiple steps, with the rate-determining step involving only NO₂ molecules.
Example mechanism: Step 1: NO₂ + NO₂ → NO₃ + NO (slow, rate-determining)
Step 2: NO₃ + CO → NO₂ + CO₂ (fast)
Overall: NO₂ + CO → NO + CO₂
This problem illustrates the connection between rate laws and reaction mechanisms. The rate law reflects the molecularity of the rate-determining step. Since the rate law is second order in NO₂ and zero order in CO, the rate-determining step must involve two NO₂ molecules and not involve CO. This shows that the overall reaction equation does not represent the actual molecular events.
Reaction Mechanism: Sequence of elementary steps
Rate-Determining Step: Slowest step in mechanism
Elementary Step: Single molecular event
• Rate law reflects rate-determining step
• Overall equation ≠ mechanism
• Elementary steps sum to overall
• Rate law reveals rate-determining step
• Look for intermediates in mechanism
• Verify steps sum to overall reaction
• Assuming overall reaction is elementary
• Not recognizing rate-determining step
• Forgetting to verify mechanism
For a reaction with the rate law Rate = k[A][B]², what are the units of the rate constant k?
Step 1: Rate = k[A][B]²
Step 2: Rate units are always M/s
Step 3: [A] has units of M, [B]² has units of M²
Step 4: M/s = k × M × M² = k × M³
Step 5: k = (M/s) / M³ = M⁻²s⁻¹
Alternatively: Units of k = M^(1-(overall order))/s
Overall order = 1 + 2 = 3
Units = M^(1-3)/s = M⁻²s⁻¹
The answer is B) M⁻²s⁻¹.
The units of the rate constant depend on the overall order of the reaction. The rate always has units of M/s, so the rate constant units must compensate for the concentration terms. For a reaction of overall order n, the rate constant has units of M^(1-n)/s. In this case, overall order = 1 + 2 = 3, so k has units of M^(1-3)/s = M⁻²s⁻¹.
Rate Constant Units: Dependent on reaction order
Dimensional Analysis: Ensuring units are consistent
Overall Order: Sum of individual reaction orders
• Rate units = M/s always
• k units = M^(1-n)/s where n = overall order
• Dimensional analysis checks consistency
• Use M^(1-n)/s formula
• Check units balance in rate law
• Zero order: k has units M/s
• Forgetting to account for all concentration terms
• Using wrong formula for k units
• Not checking dimensional consistency
Q: How do you determine reaction orders experimentally?
A: Reaction orders are determined using the method of initial rates. Multiple experiments are performed with different initial concentrations while keeping all other conditions constant. The rate is measured for each experiment, and the relationship between concentration and rate reveals the order.
For example, if doubling the concentration of A doubles the rate, the order with respect to A is 1. If doubling [A] quadruples the rate, the order is 2. This method isolates the effect of each reactant concentration on the rate, allowing determination of individual orders.
Q: How is the rate law used in industrial processes?
A: The rate law is crucial for reactor design and optimization in industrial processes:
• Residence Time: Calculating how long reactants must stay in reactor
• Temperature Control: Understanding temperature effects on rate
• Feed Ratios: Optimizing reactant concentrations based on orders
• Scale-up: Transferring lab results to industrial scale
• Safety: Predicting runaway reactions and thermal stability
Engineers use rate laws to design reactors that maximize yield while minimizing cost and ensuring safety.
Q: What's the difference between differential and integrated rate laws?
A: Differential rate laws express the rate as a function of concentration: Rate = -d[A]/dt = k[A]^n.
Integrated rate laws express concentration as a function of time: [A] = [A]₀ - kt (zero order), ln[A] = ln[A]₀ - kt (first order), or 1/[A] = 1/[A]₀ + kt (second order).
Differential laws are used to determine reaction orders experimentally. Integrated laws are used to predict concentrations at different times and calculate half-lives. Both forms are mathematically equivalent and can be derived from each other.