Reaction Rates and Chemical Equilibrium Study Notes

University of Limpopo: General Chemistry for Life & Health Sciences - Reaction Rates and Chemical Equilibrium

• Course Code: SCHE011 Section B.

• Lecturer: DR NW Masekoameng.

• Subject Focus: Chemical Kinetics (Reaction Rates) and Chemical Equilibrium.

Topic 4A: Reaction Rates - Overview and Definitions

• Chemical Kinetics Definition: The area of chemistry concerned with the speeds or rates at which a chemical reaction occurs.

• Reaction Rate Definition: The change in concentrations of reactants or products per unit of time, typically measured in Molarity per second ($M/s$).

• Observation of Rates: Reaction rates are observed when two or more reactants are put together to form one or more products.

  • General Equation: $A + B \rightarrow C$

  • Reactants ($A + B$) form Product(s) ($C$).

• Examples of Reaction Speed:

  • Slow Reaction: The rusting of iron ($Fe$). Equation: $4Fe + 3O_2 + nH_2O \rightarrow 2Fe_2O_3 \cdot nH_2O$.

  • Fast Reaction: The explosive reaction of sodium ($Na$) with water ($H_2O$).

• Importance of Kinetics:

  • Provides insights into reaction mechanisms.

  • Allows for the optimization of reaction conditions.

  • Determines the feasibility of industrial processes.

Expressing and Measuring Reaction Rates

• Rate Expressions for $A \rightarrow B$:

  • $\text{rate} = -\frac{\Delta[A]}{\Delta t}$: The rate based on the disappearance of reactant $A$. The sign is negative because the concentration of $A$ decreases over time.

  • $\text{rate} = \frac{\Delta[B]}{\Delta t}$: The rate based on the appearance of product $B$. The sign is positive because the concentration of $B$ increases over time.

• Measurement units ($\Delta [A]$ and $\Delta [B]$): Changes in molar concentration ($M$ or $\text{mol/L}$) over time interval $\Delta t$.

• Average Reaction Time: Calculated as the average time taken for product formation.

• Standard Reporting: Reaction rates are always reported as a POSITIVE value, regardless of whether the species is increasing or decreasing in concentration.

Factors Affecting Reaction Rates

• Temperature: Generally, higher temperatures lead to faster reaction rates because they increase kinetic energy, resulting in more frequent and energetic collisions.

• Concentration: Higher concentrations of reactants usually result in faster rates because they increase the frequency of particle collisions.

• Surface Area: Crushing a solid block into smaller pieces significantly increases the total surface area. More surface area exposes more reactant molecules to potential collisions.

• Collisions: A reaction occurs when particles hit each other. Only collisions with sufficient energy (Activation Energy) lead to product formation. These collisions break existing chemical bonds and allow new ones to form.

• Catalysts: Substances that change reaction rates without undergoing permanent chemical changes themselves. Enzymes act as biological catalysts in the human body.

• Pressure: For gaseous reagents, increasing pressure increases concentration (more particles in a given volume), thereby increasing the rate.

• Light: Certain reactions become more vigorous or only occur in the presence of light.

Calculation of Reaction Rates and Stoichiometry

• Units of Rate: Typically $\text{M/s}$, $\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}$, or $\text{mol} \cdot \text{dm}^{-3} \cdot \text{s}^{-1}$.

• Stoichiometric Proportionality: The rates of disappearance of reactants and appearance of products are directly proportional to their coefficients in a balanced equation.

• Example Reaction: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$.

  • For every 1 mole of $C_3H_8$ consumed, 5 moles of $O_2$ must be consumed.

  • $O_2$ reacts 5 times as fast as propane ($C_3H_8$).

  • $CO_2$ forms 3 times as fast as propane is used.

  • $H_2O$ forms 4 times as fast as propane disappears.

• Problem Scenario: $2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(g)$.

  • If Butane ($C_4H_{10}$) decreases at $0.2\,\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}$.

  • Rate of $O_2$ decrease: $\frac{13}{2} \times 0.2 = 1.3\,\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}$.

  • Rate of $CO_2$ increase: $\frac{8}{2} \times 0.2 = 0.8\,\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}$.

• Instantaneous Rate: The rate at any particular moment rather than over a long interval. It is determined by the slope ($\text{rise/run}$) of the tangent to the concentration-time curve at a specific point.

  • Equation: $\text{Slope} = \frac{[HI]_{\text{final}} - [HI]_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}$.

  • Example: A slope of $-0.027\,\text{mol/L}$ over $110\,\text{s}$ yields a rate of $2.5 \times 10^{-4}\,\text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}$.

Rate Law and Reaction Order

• Definition: A mathematical expression relating reaction rate to reactant concentration.

• General Formula: $\text{Rate} = k[A]^m[B]^n$.

  • $k$: Proportionality constant (rate constant).

  • $n, m$: Reaction orders with respect to specific reactants.

• Reaction Order properties:

  • Order is determined by the exponent on the concentration term.

  • Overall reaction order is the sum of all exponents ($m + n$).

  • Zero-order: Rate is independent of concentration.

  • First-order: If concentration doubles, rate doubles.

  • Second-order: If concentration doubles, rate quadruples ($2^2 = 4$).

• CRITICAL NOTE: Rate laws cannot be predicted from the overall balanced equation; they must be determined experimentally.

Experimental Determination of Rate Laws

• Example Data Table:

  • Exp 1: $[A]=0.10, [B]=0.10, \text{Rate}=0.20$

  • Exp 2: $[A]=0.20, [B]=0.10, \text{Rate}=0.40$

  • Exp 3: $[A]=0.30, [B]=0.10, \text{Rate}=0.60$

  • Exp 4: $[A]=0.30, [B]=0.20, \text{Rate}=2.40$

  • Exp 5: $[A]=0.30, [B]=0.30, \text{Rate}=5.40$

• Determining Order for A (m): Compare Exp 1 & 2 where $[B]$ is constant.

  • $\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.40}{0.20} = 2$

  • $\frac{[A]_2}{[A]_1} = \frac{0.20}{0.10} = 2$

  • $2 = 2^m \rightarrow m=1$ (First order in A).

• Determining Order for B (n): Compare Exp 3 & 4 where $[A]$ is constant.

  • $\frac{\text{Rate}_4}{\text{Rate}_3} = \frac{2.40}{0.60} = 4$

  • $\frac{[B]_4}{[B]_3} = \frac{0.20}{0.10} = 2$

  • $4 = 2^n \rightarrow n=2$ (Second order in B).

• Overall Order: $1 + 2 = 3$ (Third order overall).

• Determining Rate Constant ($k$): Substitute values from any experiment into the rate law.

  • Example: $0.20 = k (0.10)^1(0.10)^2 \rightarrow k = 200\,M^{-2} \cdot s^{-1}$.

• Units for $k$:

  • Order 0: $M \cdot s^{-1}$

  • Order 1: $s^{-1}$

  • Order 2: $M^{-1} \cdot s^{-1}$

  • Order 3: $M^{-2} \cdot s^{-1}$

Activation Energy and the Arrhenius Equation

• Activation Energy ($E_a$): The minimum energy required to initiate a reaction; acts as a barrier between reactants and products.

• Arrhenius Equation: $k = A e^{-E_a/RT}$.

  • $R$ (Gas Constant): $8.314\,J/K \cdot mol$.

  • $A$: Frequency factor (depends on collision frequency/orientation).

  • $T$: Temperature in Kelvin.

• Linear Form (Logarithmic): $\ln(k) = \ln(A) - \frac{E_a}{RT}$.

• Two-Temperature Formula: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.

• Example calculation:

  • $k_1 = 3.7 \times 10^{-5}/s$ at $25^{\circ}C \, (298\,K)$.

  • $k_2 = 1.8 \times 10^{-2}/s$ at $35^{\circ}C \, (308\,K)$.

  • Result: $E_a = 472\,kJ \cdot mol^{-1}$.

Reaction Mechanisms

• Reaction Mechanism: The set of elementary reactions that add up to the balanced chemical equation.

• Elementary Reaction: A single step describing the behavior of individual molecules.

• Molecularity: Number of particles colliding in an elementary step.

  • Unimolecular: 1 reactant ($CaCO_3 \rightarrow CaO + CO_2$). Order = 1.

  • Bimolecular: 2 reactants ($F + NO_2 \rightleftharpoons NO_2F$). Order = 2.

  • Termolecular: 3 reactants. Order = 3.

• Rate-Determining Step: The slowest step in a mechanism. The overall reaction rate law is determined by the rate law of this slowest step.

• Reaction Intermediate: A species produced in one step and consumed in a subsequent step. It does not appear in the overall balanced equation (e.g., $ClO$ in ozone decomposition).

• Catalyst in Mechanisms: A species present at the start, consumed in an early step, and regenerated in a later step. It does not appear in the overall balanced equation (e.g., $Cl$ in ozone decomposition).

Catalysis and Potential Energy

• Function: A catalyst increases the reaction rate by providing an alternative reaction path with a lower Activation Energy ($E_a$).

• Impact on Rate Constant ($k$): When $E_a$ is reduced, $k$ increases exponentially.

• Decomposition of Ozone Example:

  • Step 1 (Slow): $Cl(g) + O_3(g) \rightarrow ClO(g) + O_2(g)$.

  • Step 2: $ClO(g) + O(g) \rightarrow Cl(g) + O_2(g)$.

  • Overall: $O_3(g) + O(g) \rightarrow 2O_2(g)$.

  • Intermediate: $ClO$; Catalyst: $Cl$.

  • Rate Law for Overall Reaction: $\text{Rate} = k[Cl][O_3]$.

Topic 4B: Chemical Equilibrium - Introduction

• Equilibrium Definition: The condition where the forward and reverse reaction rates are equal ($\text{rate}_f = \text{rate}_r$).

• Reversible Reaction: Reactants form products, which simultaneously react back to reform reactants ($\rightleftharpoons$).

• Dynamic Equilibrium: Reactants and products are still reacting, but there is no net change in their concentrations.

• Static Equilibrium: No reaction is taking place.

• Examples of Physical Equilibrium:

  • Solubility Equilibrium: Rate of dissolution equals rate of precipitation.

  • Phase Equilibrium: Liquid-vapor equilibrium (rate of evaporation equals rate of condensation).

• Equilibrium in a Closed System: Necessary to reach equilibrium so no matter escapes.

The Equilibrium Constant ($K_c$)

• Equilibrium Law: For $aA + bB \rightleftharpoons cC + dD$, $K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$.

• Heterogeneous Equilibrium: Concentrations of pure solids ($s$) and pure liquids ($l$) are constant and are omitted (assigned a value of 1) in the $K_c$ expression.

  • Example: $H_2O(g) + C(s) \rightleftharpoons CO(g) + H_2(g)$ results in $K_c = \frac{[CO][H_2]}{[H_2O]}$.

• Manipulating $K_c$:

  • Reverse Reaction: Invert the constant ($1/K_1$).

  • Doubling Coefficients: Square the constant ($K_1^2$).

• Predicting Reaction Extent:

  • K_c > 10^2: Mostly products at equilibrium.

  • K_c < 10^{-2}: Mostly reactants at equilibrium.

  • $K_c \approx 1$: Appreciable amounts of both reactants and products.

Le Châtelier’s Principle

• Principle: If a system at equilibrium is disturbed (change in temperature, pressure, or concentration), the system shifts to counteract the change.

• Concentration Changes:

  • Add reactant: Shift forward (Right) to use it up.

  • Remove product: Shift forward (Right) to produce more.

  • Add product: Shift reverse (Left) to use it up.

• Pressure and Volume Changes (Gasses only):

  • Decrease Volume (Increase Pressure): Shifts toward the side with FEWER gas molecules.

  • Increase Volume (Decrease Pressure): Shifts toward the side with MORE gas molecules.

  • Equal moles on both sides: Pressure has no effect.

• Temperature Changes:

  • Depends on Sign of Enthalpy ($\Delta H^{\circ}$).

  • Increase Temperature (Add Heat): Shifts in the endothermic direction (absorbs heat).

  • Decrease Temperature (Remove Heat): Shifts in the exothermic direction (releases heat).

Equilibrium Calculations (ICE Table)

• Method: Use Initial, Change, and Equilibrium (ICE) amounts.

• Example: Decomposing $1.00\,mol$ of $PCl_5$ in a $1.00\,L$ container.

  • Equation: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$.

  • If $0.135\,mol$ of $PCl_3$ is present at equilibrium:

  • Initial: $PCl_5 = 1.00$, $PCl_3 = 0.0$, $Cl_2 = 0.0$.

  • Change: $PCl_5 = -0.135$, $PCl_3 = +0.135$, $Cl_2 = +0.135$.

  • Equilibrium: $PCl_5 = 0.87\,mol$, $PCl_3 = 0.135\,mol$, $Cl_2 = 0.135\,mol$.

• Finding $K_c$ from equilibrium concentrations: Substitute equilibrium molarities into the Law of Mass Action expression.

Questions & Discussion

• Q: What is the order of reaction for $BrO_3^- + 3SO_3^{2-} \rightarrow Br^- + 3SO_4^{2-}$ with rate law $\text{rate} = k[BrO_3^-][SO_3^{2-}]$?

  • A: First order with respect to $BrO_3^-$, first order with respect to $SO_3^{2-}$, and second order overall ($1+1=2$).

• Q: Describe the feasibility of fixing nitrogen via $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ at $25^{\circ}C$ given $K_c = 1 \times 10^{-30}$.

  • A: Extremely poor choice. Since $K_c$ is very small, the equilibrium lies far to the left (reactants), and very little $NO$ is produced.

• Q: Predicted rate law for decomposition of $H_2O_2$ catalyzed by $I^-$ where the first step ($H_2O_2 + I^- \rightarrow H_2O + IO^-$) is slow?

  • A: $\text{Rate} = k[H_2O_2][I^-]$. Here, $I^-$ is a catalyst and $IO^-$ is an intermediate.