Introduction#
Consider a general elementary reaction: $$ a \mathrm{A} + b \mathrm{B} \rightarrow c \mathrm{C} + d \mathrm{D} $$ By convention, since we would like the rate to be positive if the reaction proceeds from left to right, we choose positive derivatives for the products and negative ones for the reactants. The differential rate law can be written as: $$ \frac{1}{c} \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{d}t} = \frac{1}{d} \frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t} = -\frac{1}{a} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -\frac{1}{b} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = k [\mathrm{A}]^{a} [\mathrm{B}]^b $$ The order of the reaction is the sum of the exponents of the reactants in the rate law. The overall order is the sum of the orders with respect to each reactant. The rate constant \(k\) is temperature-dependent and is independent of the concentrations of the reactants.
The most common elementary reactions are zero-order, first-order, and second-order reactions, while third-order and higher-order reactions are rare. The integrated rate laws for these reactions are derived below.
Zero-Order Reaction#
For a zero-order reaction, the rate law is given by: $$ \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k $$ Integrating this equation gives: $$ [\mathrm{A}] = -kt + [\mathrm{A}(0)] $$ This equation describes the concentration of the reactant as a function of time.
First-Order Reaction#
For a first-order reaction, the rate law is given by: $$ \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k[\mathrm{A}] $$ Integrating this equation gives: $$ \ln [\mathrm{A}] = -kt + \ln [\mathrm{A}(0)] $$ This equation describes the concentration of the reactant as a function of time.
Second-Order Reaction#
\(2 \mathrm{A} \rightarrow \mathrm{products}\)#
For the reaction \(2 \mathrm{A} \rightarrow \mathrm{products}\), the differential rate law is given by: $$ \frac{1}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = k[\mathrm{A}]^2 $$ Let \(x\) represent the extent of reaction, where \(x\) is the change in concentration of \(\mathrm{A}\). Then: $$ [\mathrm{A}] = [\mathrm{A}(0)] - 2x $$ Substituting into the rate law gives: $$ \frac{\mathrm{d}x}{\mathrm{d}t} = k ([\mathrm{A}(0)] - 2x)^2 $$ Separating variables: $$ \int \frac{\mathrm{d}x}{([\mathrm{A}(0)] - 2x)^2} = k \int \mathrm{d}t $$ Integrating both sides: $$ -\frac{1}{2[\mathrm{A}]} + \frac{1}{2[\mathrm{A}(0)]} = kt $$ Rearranging: $$ \frac{1}{[\mathrm{A}]} = 2kt + \frac{1}{[\mathrm{A}(0)]} $$ This equation describes the concentration of the reactant as a function of time.
\(\mathrm{A} + \mathrm{B} \rightarrow \mathrm{products}\)#
For the reaction \(\mathrm{A} + \mathrm{B} \rightarrow \mathrm{products}\), the differential rate law is given by: $$ -\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = k[\mathrm{A}][\mathrm{B}] $$ Let \(x\) represent the extent of reaction. Then: $$ [\mathrm{A}] = [\mathrm{A}(0)] - x, \quad [\mathrm{B}] = [\mathrm{B}(0)] - x $$ Substituting into the rate law: $$ \frac{\mathrm{d}x}{\mathrm{d}t} = k([\mathrm{A}(0)] - x)([\mathrm{B}(0)] - x) $$ Separating variables: $$ \int \frac{\mathrm{d}x}{([\mathrm{A}(0)] - x)([\mathrm{B}(0)] - x)} = k \int \mathrm{d}t $$ Using partial fraction decomposition: $$ \frac{1}{([\mathrm{A}(0)] - x)([\mathrm{B}(0)] - x)} = \frac{1}{[\mathrm{A}(0)] - [\mathrm{B}(0)]} \left(\frac{1}{[\mathrm{B}(0)] - x} - \frac{1}{[\mathrm{A}(0)] - x}\right) $$ Integrating both sides: $$ \frac{1}{[\mathrm{A}(0)] - [\mathrm{B}(0)]} \ln \frac{[\mathrm{A}][\mathrm{B}(0)]}{[\mathrm{A}(0)][\mathrm{B}]} = kt $$
It often occurs for second-order reactions that the rate law involves two reactants, one of which is in large excess. In this case, the rate law can be simplified to pseudo-first-order: $$ \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k[\mathrm{A}][\mathrm{B}(0)] $$ Integrating this equation gives: $$ \ln \left[\frac{[\mathrm{A}(t)]}{[\mathrm{A}(0)]}\right] = -k[\mathrm{B}(0)]t $$
Third-Order Reaction#
\(\text{A}+\text{B}+\text{C} \rightarrow \text{products}\)#
For a third-order reaction involving three different reactants, the differential rate law is: $$ -\frac{d[\text{A}]}{dt} = -\frac{d[\text{B}]}{dt} = -\frac{d[\text{C}]}{dt} = k[\text{A}][\text{B}][\text{C}] $$
Let \(x\) represent the extent of reaction. Then: $$ [\text{A}] = [\text{A}(0)] - x, \quad [\text{B}] = [\text{B}(0)] - x, \quad [\text{C}] = [\text{C}(0)] - x $$
Substituting into the rate law: $$ \frac{dx}{dt} = k([\text{A}(0)] - x)([\text{B}(0)] - x)([\text{C}(0)] - x) $$
Separating Variables:
$$ \int \frac{dx}{([\text{A}(0)] - x)([\text{B}(0)] - x)([\text{C}(0)] - x)} = k \int dt $$
Using partial fraction decomposition:
$$ \frac{1}{([\text{A}(0)] - x)([\text{B}(0)] - x)([\text{C}(0)] - x)} = \frac{A}{[\text{A}(0)] - x} + \frac{B}{[\text{B}(0)] - x} + \frac{C}{[\text{C}(0)] - x} $$
Multiplying both sides by \(([\text{A}(0)] - x)([\text{B}(0)] - x)([\text{C}(0)] - x)\) and expanding, we find the coefficients \(A\), \(B\), and \(C\) by matching terms. Solving the resulting system:
$$ A = \frac{1}{([\text{B}(0)] - [\text{A}(0)])([\text{C}(0)] - [\text{A}(0)])}, \quad B = \frac{1}{([\text{A}(0)] - [\text{B}(0)])([\text{C}(0)] - [\text{B}(0)])}, \quad C = \frac{1}{([\text{A}(0)] - [\text{C}(0)])([\text{B}(0)] - [\text{C}(0)])} $$
Thus, the integral becomes
$$ A \int \frac{dx}{[\text{A}(0)] - x} + B \int \frac{dx}{[\text{B}(0)] - x} + C \int \frac{dx}{[\text{C}(0)] - x} = kt + \text{constant} $$
Integrating both sides, combining the constants, the integrated rate law is:
$$ \frac{1}{([\text{B}(0)] - [\text{A}(0)])([\text{C}(0)] - [\text{A}(0)])} \ln\left(\frac{[\text{B}][\text{C}][\text{A}(0)]}{[\text{B}(0)][\text{C}(0)][\text{A}]}\right) + \frac{1}{([\text{A}(0)] - [\text{B}(0)])([\text{C}(0)] - [\text{B}(0)])} \ln\left(\frac{[\text{A}][\text{C}][\text{B}(0)]}{[\text{A}(0)][\text{C}(0)][\text{B}]}\right) + \frac{1}{([\text{A}(0)] - [\text{C}(0)])([\text{B}(0)] - [\text{C}(0)])} \ln\left(\frac{[\text{A}][\text{B}][\text{C}(0)]}{[\text{A}(0)][\text{B}(0)][\text{C}]}\right) = kt $$
\(2 \text{A} + \text{B} \rightarrow \text{products}\)#
We start with the differential rate law:
$$ \frac{1}{2} \frac{d[\text{A}]}{dt} = -\frac{d[\text{B}]}{dt} = k[\text{A}]^2[\text{B}] $$
Let \(x\) represent the extent of reaction, where \(x\) is the change in concentration of \(\text{B}\). Then
$$ [\text{A}] = [\text{A}(0)] - 2x, \quad [\text{B}] = [\text{B}(0)] - x $$
Substituting into the rate law gives
$$ \frac{dx}{dt} = k ([\text{A}(0)] - 2x)^2 ([\text{B}(0)] - x) $$
Separate variables:
$$ \int \frac{dx}{([\text{A}(0)] - 2x)^2 ([\text{B}(0)] - x)} = k \int dt $$
Using partial fraction decomposition,
$$ \frac{1}{([\text{A}(0)] - 2x)^2 ([\text{B}(0)] - x)} = \frac{A}{[\text{A}(0)] - 2x} + \frac{B}{([\text{A}(0)] - 2x)^2} + \frac{C}{[\text{B}(0)] - x} $$
Multiplying both sides by \(([\text{A}(0)] - 2x)^2 ([\text{B}(0)] - x)\) and expanding, we find the coefficients \(A\), \(B\), and \(C\) by matching terms. Solving the resulting system,
$$ A = -\frac{2}{\left( [\text{A}(0)] - 2[\text{B}(0)] \right)^2}, \quad B = -\frac{2}{[\text{A}(0)] - 2[\text{B}(0)]}, \quad C = \frac{1}{\left( [\text{A}(0)] - 2[\text{B}(0)] \right)^2} $$
Thus, the integral becomes
$$ \int \frac{dx}{([\text{A}(0)] - 2x)^2 ([\text{B}(0)] - x)} = -\frac{1}{\left( [\text{A}(0)] - 2[\text{B}(0)] \right)} \left( \frac{1}{[\text{A}]} - \frac{1}{[\text{A}(0)]} \right) - \frac{1}{\left( [\text{A}(0)] - 2[\text{B}(0)] \right)^2} \ln \frac{[\text{A}][\text{B}(0)]}{[\text{A}(0)][\text{B}]} $$
Integrating both sides, combining the constants, the integrated rate law is:
$$ \frac{1}{[\text{A}(0)] - 2[\text{B}(0)]} \left( \frac{1}{[\text{A}(0)]} - \frac{1}{[\text{A}]} \right) + \frac{1}{\left( [\text{A}(0)] - 2[\text{B}(0)] \right)^2} \ln \frac{[\text{A}][\text{B}(0)]}{[\text{A}(0)][\text{B}]} = kt $$
Basic Reaction Mechanisms#
Opposing Reactions#
Consider the opposing first-order reaction \(\text{A} \rightleftharpoons \text{B}\) with rate constants \(k_1\) in the forward direction and \(k_{-1}\) in the reverse direction. The differential rate law is given by:
$$ \frac{dx}{dt} = k_1 [\text{A}] - k_{-1} [\text{B}] $$
We can rearrange the differential equation to isolate \( \frac{\mathrm{d}x}{\mathrm{d}t} \):
$$ \frac{\mathrm{d}x}{\mathrm{d}t} + (k_1 + k_{-1})x = k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)] $$
This is a linear differential equation of the form:
$$ \frac{\mathrm{d}x}{\mathrm{d}t} + P x = Q $$
where:
- \( P = k_1 + k_{-1} \)
- \( Q = k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)] \)
To solve this, we use an integrating factor \( \mu(t) \):
$$ \mu(t) = e^{\int P \mathrm{d}t} = e^{(k_1 + k_{-1})t} $$
Multiplying both sides of the differential equation by \( \mu(t) \):
$$ e^{(k_1 + k_{-1})t} \frac{\mathrm{d}x}{\mathrm{d}t} + (k_1 + k_{-1}) e^{(k_1 + k_{-1})t} x = (k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]) e^{(k_1 + k_{-1})t} $$
The left side simplifies to the derivative of \( x \mu(t) \):
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left( x e^{(k_1 + k_{-1})t} \right) = (k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]) e^{(k_1 + k_{-1})t} $$
Integrating both sides with respect to \( t \):
$$ x e^{(k_1 + k_{-1})t} = \int (k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]) e^{(k_1 + k_{-1})t} \mathrm{d}t + C $$
Performing the integration:
$$ x e^{(k_1 + k_{-1})t} = \frac{k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]}{k_1 + k_{-1}} e^{(k_1 + k_{-1})t} + C $$
Solving for \( x \):
$$ x = \frac{k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]}{k_1 + k_{-1}} + C e^{-(k_1 + k_{-1})t} $$
Applying the initial condition \( x(0) = 0 \):
$$ 0 = \frac{k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]}{k_1 + k_{-1}} + C $$
Thus, \( C = -\frac{k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]}{k_1 + k_{-1}} \), and the solution becomes:
$$ x(t) = \frac{k_1 [\text{A}(0)] - k_{-1} [\text{B}(0)]}{k_1 + k_{-1}} \left( 1 - e^{-(k_1 + k_{-1})t} \right) $$
Finally, the concentrations as functions of time are:
$$ [\text{A}] = [\text{A}(0)] - x(t) $$
$$ [\text{B}] = [\text{B}(0)] + x(t) $$
Parallel Reactions#
Consider two competing first-order parallel reactions:
$$ \text{A} \xrightarrow{k_1} \text{B} $$
$$ \text{A} \xrightarrow{k_2} \text{C} $$
The differential rate laws are:
$$ \frac{d[\text{B}]}{dt} = k_1 [\text{A}] $$
$$ \frac{d[\text{C}]}{dt} = k_2 [\text{A}] $$
The total rate of disappearance of \(\text{A}\) is the sum of the rates of the two reactions:
$$ \frac{d[\text{A}]}{dt} = -\frac{d[\text{B}]}{dt} - \frac{d[\text{C}]}{dt} = -(k_1 + k_2) [\text{A}] $$
Integrating this equation gives:
$$ \ln \frac{[\text{A}]}{[\text{A}(0)]} = -(k_1 + k_2) t $$
$$ \frac{d[\text{B}]}{dt} = k_1 [\text{A}(0)] e^{-(k_1 + k_2) t} $$
$$ \frac{d[\text{C}]}{dt} = k_2 [\text{A}(0)] e^{-(k_1 + k_2) t} $$
$$ [\text{B}(t)] = \int k_1 [\text{A}(0)] e^{-(k_1 + k_2) t} dt = \frac{k_1 [\text{A}(0)]}{k_1 + k_2} (1 - e^{-(k_1 + k_2) t}) $$
$$ [\text{C}(t)] = \int k_2 [\text{A}(0)] e^{-(k_1 + k_2) t} dt = \frac{k_2 [\text{A}(0)]}{k_1 + k_2} (1 - e^{-(k_1 + k_2) t}) $$
Consecutive Reactions#
Consider two consecutive first-order reactions:
$$ \text{A} \xrightarrow{k_1} \text{B} \xrightarrow{k_2} \text{C} $$
The differential rate laws are:
$$ -\frac{d[\text{A}]}{dt} = k_1 [\text{A}] $$
$$ \frac{d[\text{B}]}{dt} = k_1 [\text{A}] - k_2 [\text{B}] $$
$$ \frac{d[\text{C}]}{dt} = k_2 [\text{B}] $$
Integrating the first equation gives:
$$ [\text{A}] = [\text{A}(0)] e^{-k_1 t} $$
To solve for \( [\mathrm{B}(t)] \), we start with the differential rate law:
$$ \frac{d[\text{B}]}{dt} = k_1 [\text{A}] - k_2 [\text{B}] $$
Given that \([\text{A}] = [\text{A}(0)] e^{-k_1 t}\), substitute this into the equation:
$$ \frac{d[\text{B}]}{dt} + k_2 [\text{B}] = k_1 [\text{A}(0)] e^{-k_1 t} $$
This is a linear differential equation of the form:
$$ \frac{d[\text{B}]}{dt} + P [\text{B}] = Q(t) $$
where:
- \( P = k_2 \)
- \( Q(t) = k_1 [\text{A}(0)] e^{-k_1 t} \)
The integrating factor \(\mu(t)\) is:
$$ \mu(t) = e^{\int P dt} = e^{k_2 t} $$
Multiply both sides of the differential equation by the integrating factor:
$$ e^{k_2 t} \frac{d[\text{B}]}{dt} + k_2 e^{k_2 t} [\text{B}] = k_1 [\text{A}(0)] e^{-(k_1 - k_2) t} $$
The left side becomes the derivative of \([\text{B}] e^{k_2 t}\):
$$ \frac{d}{dt} \left( [\text{B}] e^{k_2 t} \right) = k_1 [\text{A}(0)] e^{-(k_1 - k_2) t} $$
Integrate both sides with respect to \(t\):
$$ [\text{B}] e^{k_2 t} = \int k_1 [\text{A}(0)] e^{-(k_1 - k_2) t} dt + C $$
Perform the integration:
$$ [\text{B}] e^{k_2 t} = \frac{k_1 [\text{A}(0)]}{k_2 - k_1} e^{-(k_1 - k_2) t} + C $$
Solve for \([\text{B}]\):
$$ [\text{B}] = \frac{k_1 [\text{A}(0)]}{k_2 - k_1} \left( e^{-k_1 t} - e^{-k_2 t} \right) $$
Finally, since by mass conservation \( [\text{A}] + [\text{B}] + [\text{C}] = [\text{A}(0)] \), we have:
$$ [\text{C}] = [\text{A}(0)] - [\text{A}] - [\text{B}] = [\text{A}(0)] (1 - e^{-k_1 t}) - \frac{k_1 [\text{A}(0)]}{k_2 - k_1} (e^{-k_1 t} - e^{-k_2 t}) $$
If \( k_1 \gg k_2 \), the rate of formation of \( \text{B} \) is approximately:
$$ \frac{d[\text{B}]}{dt} = [\text{A}(0)] (e^{-k_2 t}-e^{-k_1 t}) $$
If \( k_2 \gg k_1 \), the rate of formation of \(\text{B}\) is approximately:
$$ \frac{d[\text{B}]}{dt} = \frac{k_1 [\text{A}(0)]}{k_2} (e^{-k_2 t}-e^{-k_1 t}) $$
Steady-State Approximation#
Unimolecular Decomposition: The Lindemann-Hinshelwood Mechanism#
The Lindemann-Hinshelwood mechanism provides a detailed explanation for unimolecular decomposition reactions through a two-step process involving molecular collisions. The mechanism consists of the following elementary steps:
Activation Step (Collision): $$ \text{A} + \text{M} \rightleftharpoons \text{A}^* + \text{M}, \quad \text{rate constants} \quad k_1, \, k_{-1} $$
Decomposition Step: $$ \text{A}^* \rightarrow \text{Products}, \quad \text{rate constant} \quad k_2 $$
Here, \(\text{A}\) is the reactant molecule, \(\text{M}\) is a third body that facilitates the energy transfer, and \(\text{A}^*\) is the energized intermediate.
Differential Rate Laws:
Let \([\text{A}]\) represent the concentration of reactant \(\text{A}\), \([\text{M}]\) the concentration of the third body, and \([\text{A}^*]\) the concentration of the energized intermediate. The differential rate laws for the system are:
$$ \frac{d[\text{A}]}{dt} = -k_1 [\text{A}][\text{M}] + k_{-1} [\text{A}^*][\text{M}] $$
$$ \frac{d[\text{A}^* ]}{dt} = k_1 [\text{A}][\text{M}] - k_{-1} [\text{A}^* ][\text{M}] - k_2 [\text{A}^*] $$
Assuming the steady-state approximation for the intermediate \(\text{A}^* \) (i.e., \(\frac{d[\text{A}^*]}{dt} \approx 0\)), we set:
$$ k_1 [\text{A}][\text{M}] = (k_{-1} [\text{M}] + k_2) [\text{A}^*] $$
Solving for \([\text{A}^*]\):
$$ [\text{A}^*] = \frac{k_1 [\text{A}][\text{M}]}{k_{-1} [\text{M}] + k_2} $$
Integrated Rate Law:
The rate of product formation is given by:
$$ \frac{d[\text{Products}]}{dt} = k_2 [\text{A}^*] = \frac{k_1 k_2 [\text{A}][\text{M}]}{k_{-1} [\text{M}] + k_2} $$
If the concentration of \(\text{M}\) remains constant (i.e., \(\text{M}\) is in large excess), we can define an effective rate constant:
$$ k_{\text{eff}} = \frac{k_1 k_2 [\text{M}]}{k_{-1} [\text{M}] + k_2} $$
Thus, the rate law simplifies to pseudo-first-order:
$$ \frac{d[\text{A}]}{dt} = -k_{\text{eff}} [\text{A}] $$
Integrating this equation yields:
$$ [\text{A}] = [\text{A}(0)] e^{-k_{\text{eff}} t} $$
Therefore, the concentration of reactant \(\text{A}\) decreases exponentially with time under the steady-state approximation.
Enzyme-Catalyzed Reactions: Michaelis-Menten Mechanism#
The Michaelis-Menten mechanism provides a fundamental framework for understanding enzyme-catalyzed reactions. It describes how enzymes interact with substrates to form products, highlighting the relationship between enzyme concentration, substrate concentration, and reaction rate. Consider the enzymatic reaction:
$$ \text{E} + \text{S} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \text{ES} \xrightarrow{k_2} \text{E} + \text{P} $$
where \( \text{E} \) is the enzyme, \( \text{S} \) is the substrate, \( \text{ES} \) is the enzyme-substrate complex, and \( \text{P} \) is the product. The rate constants are \( k_1 \) for the formation of \( \text{ES} \), \( k_{-1} \) for the dissociation of \( \text{ES} \), and \( k_2 \) for the conversion of \( \text{ES} \) to product. The rate of change of the enzyme-substrate complex is given by the differential equation:
$$ \frac{d[\text{ES}]}{dt} = k_1 [\text{E}][\text{S}] - (k_{-1} + k_2) [\text{ES}] $$
Assuming the steady-state approximation where \( \frac{d[\text{ES}]}{dt} \approx 0 \), the equation simplifies to:
$$ k_1 [\text{E}][\text{S}] = (k_{-1} + k_2) [\text{ES}] $$
Solving for \( [\text{ES}] \) yields:
$$ [\text{ES}] = \frac{k_1 [\text{E}][\text{S}]}{k_{-1} + k_2} $$
Defining the Michaelis constant \( K_m = \frac{k_{-1} + k_2}{k_1} \), we can rewrite \( [\text{ES}] \) as:
$$ [\text{ES}] = \frac{[\text{E}][\text{S}]}{K_m} $$
Since the total enzyme concentration is \( [\text{E}]{\text{total}} = [\text{E}] + [\text{ES}] \), we can express \( [\text{E}] \) as \( [\text{E}] = [\text{E}]{\text{total}} - [\text{ES}] \). Substituting back, we get:
$$ [\text{ES}] = \frac{([\text{E}]_{\text{total}} - [\text{ES}])[\text{S}]}{K_m} $$
Rearranging, we find:
$$ [\text{ES}] = \frac{[\text{E}]_{\text{total}}[\text{S}]}{K_m + [\text{S}]} $$
The rate of product formation is then:
$$ \frac{d[\text{P}]}{dt} = k_2 [\text{ES}] = \frac{k_2 [\text{E}]_{\text{total}} [\text{S}]}{K_m + [\text{S}]} $$
Defining \( V_{\max} = k_2 [\text{E}]_{\text{total}} \), the equation becomes:
$$ \frac{d[\text{P}]}{dt} = \frac{V_{\max} [\text{S}]}{K_m + [\text{S}]} $$
This is the Michaelis-Menten equation, which describes how the reaction rate depends on substrate concentration. At low substrate concentrations (\( [\text{S}] \ll K_m \)), the rate is proportional to \( [\text{S}] \), while at high concentrations (\( [\text{S}] \gg K_m \)), the rate approaches \( V_{\max} \). Inhibitors can affect enzyme activity by interacting with the enzyme or enzyme-substrate complex. Competitive inhibitors compete with the substrate for binding to the active site, increasing \( K_m \) without affecting \( V_{\max} \). The modified rate equation is:
$$ \frac{d[\text{P}]}{dt} = \frac{V_{\max} [\text{S}]}{K_m \left(1 + \frac{[\text{I}]}{K_i}\right) + [\text{S}]} $$
where \( [\text{I}] \) is the inhibitor concentration and \( K_i \) is the inhibition constant. Non-competitive inhibitors bind to an allosteric site, decreasing \( V_{\max} \) without changing \( K_m \). The rate equation becomes:
$$ \frac{d[\text{P}]}{dt} = \frac{\left( \frac{V_{\max}}{1 + \frac{[\text{I}]}{K_i}} \right) [\text{S}]}{K_m + [\text{S}]} $$
Uncompetitive inhibitors bind only to the enzyme-substrate complex, reducing both \( V_{\max} \) and \( K_m \). The equation is modified to:
$$ \frac{d[\text{P}]}{dt} = \frac{V_{\max} [\text{S}]}{\left( K_m + [\text{S}] \right) \left( 1 + \frac{[\text{I}]}{K_i} \right)} $$
Understanding these inhibition mechanisms is crucial for designing effective drugs and regulating biochemical pathways.