Rate Law and Specific Rate Constant
1.0Rate of a Reaction
The rate of a reaction measures the change in the concentration of reactants or products per unit time. For a general reaction aA+bB→cC+dD, the rate can be expressed as:
Rate=−a1ΔtΔ[A]=−b1ΔtΔ[B]=+c1ΔtΔ[C]=+d1ΔtΔ[D]
The negative sign indicates a decrease in concentration of reactants over time.
2.0Rate Law and Specific Rate Constant
3.0Rate Law
The rate law, also known as the rate equation, is an experimentally determined expression that links the rate of a chemical reaction to the concentrations of the reactants. It is a fundamental part of chemical kinetics.
For a general reaction: aA+bB→Products
The rate law formula is given by:
Rate=k[A]x[B]y
- [A] and [B] are the molar concentrations of reactants A and B.
- k is the specific rate constant.
- x and y are the orders of the reaction with respect to reactants A and B. They are determined experimentally and need not equal the stoichiometric coefficients a and b.
Specific Rate Constant (k)
The specific rate constant (k) is the proportionality constant in the rate law. It represents the rate of the reaction when the concentration of each reactant is unity (1 M). The value of k is unique for a given reaction at a specific temperature.
- Significance: A large value of k indicates a fast reaction, while a small value signifies a slow one.
- Temperature Dependence: The value of k is highly dependent on temperature, as described by the Arrhenius equation. As the temperature increases, the value of k generally increases.
- Units: The units of the specific rate constant depend on the overall order of the reaction.
Rate Law and Specific Rate Constant Formula
- Rate Law Formula:
Rate=k[A]x[B]y
- Order of Reaction:
The order of reaction is the sum of the powers of the concentration terms in the rate law.
Order=x+y
- Unit of Specific Rate Constant (k):
Depends on the order of reaction.
For a first-order reaction: k=s−1
For a second-order reaction: k=mol−1⋅L⋅s−1
General: [k]=(mol L−1)1−ns−1
where n = order of reaction.
4.0Difference between Rate Law and Specific Rate Constant
5.0Rate Law and Specific Rate Constant Practice Problem
Practice Problem
Problem: For the reaction 2\text{NO}(\text{g}) + 2\text{H}_2(\text{g}) \longrightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g}) , the rate law is found to be \text{Rate} = k[\text{NO}]^2 [\text{H}_2]
(a) What is the overall order of the reaction?
(b) What are the units of the specific rate constant (k)?
(c) How does the rate change if the concentration of NO is doubled and the concentration of H_2 is halved?
Solution:
(a) Overall Order: The exponents in the rate law are 2 for NO and 1 for H_2. Overall Order = 2+1=3. The reaction is third order.
(b) Units of k: For a third-order reaction (n=3), the units are:
[k]=(mol L−1)1−3s−1=(mol L−1)−2s−1=L2mol−2s−1
(c) Change in Rate:
Original RateNew Rate=k[NO]2[H2]=k[2NO]2[2H2]=k(4)[NO]2(21)[H2]=2k[NO]2[H2]
The new rate is twice the original rate.