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# Chapter 13 Chemical Equilibrium

Shaun Williams, PhD

### Reaction Rate

• Change in concentration of a reactant or product per unit time.
• $$\textrm{Rate}=\frac{\textrm{conc. of A at }t_2 - \textrm{conc. of A at }t_1}{t_2 - t_1}$$ $$\textrm{Rate}=\frac{\Delta [\textrm{A}]}{\Delta t}$$
• $$[\mathrm{A}]$$ means concentration of A in mol/L
• A is the reactant or product being considered

### The Decomposition of Nitrogen Dioxide $$\chem{2NO_2(g) \rightarrow 2NO(g) + O_2(g)}$$ (at $$300^\circ \textrm{C}$$)

Time
(sec)
$$[\mathrm{NO}_2]$$
(mol/L)
$$[\mathrm{NO}]$$
(mol/L)
$$[\mathrm{O}_2]$$
(mol/L)
0 0.0100 0 0
50 0.0079 0.0021 0.0011
100 0.0065 0.0035 0.0018
150 0.0055 0.0045 0.0023
200 0.0048 0.0052 0.0026
250 0.0043 0.0057 0.0029
350 0.0034 0.0066 0.0033
400 0.0031 0.0069 0.0035

### Instantaneous Rate

• Value of the rate at a particular time.
• Can be obtained by computing the slope of a line tangent to the curve at that point.

### Rate Law

• Shows how the rate depends on the concentrations of reactants.
• For the decomposition of nitrogen dioxide: $$\chem{2NO_2(g)\rightarrow 2NO(g)+O_2(g)}$$ $$\mathrm{Rate}=k\conc{NO_2}^2$$
• $$k$$ is the rate constant
• $$n$$ is the order of the reactant
• The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.
• The value of the exponent $$n$$ must be determined by experiment; it cannot be written from the balanced equation.

### Types of Rate Laws

• Differential Rate Law (rate law) - shows how the rate of a reaction depends on concentrations.
• Integrated Rate Law - shows how the concentrations of species in the reaction depend on time.

### Rate Laws: A Summary

• Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants.
• Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient.
• Experimental convenience usually dictates which type of rate law is determined experimentally.
• Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

### What does it mean to determine the form of the rate law?

• Determine experimentally the power to which each reactant concentration must be raised in the rate law.

### Method of Initial Rates

• The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible.
• Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run.
• The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants.

### Overall Reaction Order

• The sum of the exponents in the reaction rate equation. $$\mathrm{Rate}=k\conc{A}^n\conc{B}^m$$ $$\text{Overall reaction order} = n+m$$
• $$k$$ is the rate constant
• $$\conc{A}$$ is the concentration of reactant A
• $$\conc{B}$$ is the concentration of reactant B

### First-Order

• $$\mathrm{Rate}=k\conc{A}$$
• Integrated: $$\ln\conc{A}=-kt+\ln\conc{A}_0$$
• $$\conc{A}$$ is the concentration of A at time t
• $$k$$ is the rate constant
• $$t$$ is the time
• $$\conc{A}_0$$ is the initial concentration of A

### Plot of $$\ln\conc{N_2O_5}$$ versus Time

$$\ln\conc{N_2O_5}$$ Time(s)
-2.303 0
-2.649 50
-2.996 100
-3.689 200
-4.382 300
-5.075 400

### Half-Life in First-Order Reactions

• Time required for a reactant to reach half its original concentration
• Half-life: $$t_\bfrac{1}{2} = \frac{0.693}{k}$$ where $$k$$ is the rate constant
• Half-life does not depend on the concentration of reactants.

### Example Problem

A first order reaction is 35% complete at teh end of 55 minutes. What is the value of $$k$$?

$$k=7.8 \times 10^{-3}\,\mathrm{min}^{-1}$$

### Second-Order Reactions

• $$\mathrm{Rate}=k\conc{A}^2$$
• Integrated: $$\frac{1}{\conc{A}} = kt+\frac{1}{\conc{A}_0}$$
• $$\conc{A}$$ is the concentration of A at time t
• $$k$$ is the rate constant
• $$t$$ is the time
• $$\conc{A}_0$$ is the initial concentration of A

### Half-Life in Second-Order Reactions

• Half-Life: $$t_\bfrac{1}{2}=\frac{1}{k\conc{A}_0}$$
• $$k$$ is the rate constant
• $$\conc{A}_0$$ is the initial concentration of A
• Half-life gets longer as the reaction progresses and the concentration of reactants decrease.
• Each successive half-life is double the preceding one.

### Practice Problem

For a reaction $$\chem{aA\rightarrow Products}$$, $$\conc{A}_0=5.0\,\mathrm{M}$$, and the first two half-lives are 25 and 50 minutes, respectively.

1. Write the rate law for this reaction

$$\mathrm{rate}=k\conc{A}^2$$

2. Calculate $$k$$.

$$k=8.0 \times 10^{-3}\,\mathrm{M}^{-1}\,\mathrm{min}^{-1}$$

3. Calculate $$\conc{A}$$ at $$t=525\,\mathrm{minutes}$$.

$$\conc{A}=0.23\,\mathrm{M}$$

### Zero-Order Reactions

• $$\mathrm{Rate}=k\conc{A}^0=k$$
• Integrated: $$\conc{A}=-kt+\conc{A}_0$$
• $$\conc{A}$$ is the concentration of A at time t
• $$k$$ is the rate constant
• $$t$$ is the time
• $$\conc{A}_0$$ is the initial concentration of A

### Half-Life in a Zero-Order Reaction

• Half-Life: $$t_\bfrac{1}{2} = \frac{\conc{A}_0}{2k}$$
• $$k$$ is the rate constant
• $$\conc{A}_0$$ is the intial concentration of A
• Half-life gets shorter as the reaction progresses and the concentration of reactants decrease.

### Reaction Mechanism

• Most chemical reactions occur by a series of elementary steps.
• An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction.

### A Molecular Representation of the Elementary Steps in the Reaction of $$\chem{NO_2}$$ and $$\chem{CO}$$

$$\chem{NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)}$$

### Elementary Steps (Molecularity)

• Unimolecular - reaction involving one molecule; first order
• Bimolecular - reaction involving the collision of two species; second order
• Termolecular - reaction involving the collision of three speciesl third order
• Very rare!

### Rate-Determining Step

• A reaction is only as fast as its slowest step.
• The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction.

### Reaction Mechanism Requirements

• The sum of the elementary steps must give the overall balanced equation for the reaction.
• The mechanism must agree with the experimentally determined rate law.

### Decomposition of $$\chem{N_2O_5}$$

$$\chem{2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)}$$

\begin{align} \text{Step 1:} & \; \chem{2N_2O_5 \rightleftharpoons 2NO_2+2NO_3} \; & \text{(fast)} \\ \text{Step 2:} & \; \chem{NO_2+NO_3 \rightarrow NO+O_2+NO_2} \; & \text{(slow)} \\ \text{Step 3:} & \; \chem{NO_3+NO \rightarrow 2NO_2} \; & \text{(fast)} \end{align}

### Collision Model

• Molecules must collide to react
• Main Factors:
• Activation energy, $$E_a$$
• Temperature
• Molecular orientations
• Activation Energy, $$E_a$$, is the energy that must be overcome to produce a chemical reaction

### For Reactants to Form Products

• Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy).
• Relative orientation of the reactants must allow formation of any new bonds necessary to produce products.

### Arrhenius Equation

$$k=Ae^{-\bfrac{E_a}{RT}}$$

• $$A$$ is a frequency factor
• $$E_a$$ is the activation energy
• $$R$$ is the gas constant ($$8.3145\,\bfrac{\mathrm{J}}{\mathrm{K\cdot mol}}$$)
• $$T$$ is the temperature in Kelvin

Linear form: $$\ln(k)=-\frac{E_a}{R}\left(\frac{1}{T}\right)+\ln(A)$$

### Catalyst

• A substance that speeds up a reaction without being consumed itself.
• Provides a new pathway for the reaction with a lower activation energy.

### Heterogeneous Catalysis

• Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst.
• Adsorption – collection of one substance on the surface of another substance.
• Steps in heterogeneous catalysis:
1. Adsorption and activation of the reactants.
2. Migration of the adsorbed reactants on the surface.
3. Reaction of the adsorbed substances.
4. Escape, or desorption, of the products.
• Homogeneous catalysis
• Exists in the same phase as the reacting molecules.
• Enzymes are nature’s catalysts.

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