Chemistry - Reaction Rates

What are Reaction Rates?

A reaction rate is simply a measure of how fast a chemical reaction occurs. It describes how quickly the reactants are used up or how quickly the products are formed over a specific period of time. You can think of it like the speed of a car, but instead of measuring miles per hour, chemists measure change in concentration per second or minute.

Determining Rate Laws from Experimental Data

To find the rate law, we look at how the initial rate of a reaction changes when we change the concentration of one reactant at a time. 

Rate Law Format: Rate = k[A]^x[B]^y

k = Rate constant

x = Order of reactant A

y = Order of reactant B

Practice Problem 1: Determining Reaction Orders

Consider the following reaction: 2 NO(g) + Cl2(g) -> 2 NOCl(g)

The following data was collected at a constant temperature:

Trial 1: [NO] = 0.10 M, [Cl2] = 0.10 M, Initial Rate = 0.18 M/s

Trial 2: [NO] = 0.10 M, [Cl2] = 0.20 M, Initial Rate = 0.36 M/s

Trial 3: [NO] = 0.20 M, [Cl2] = 0.10 M, Initial Rate = 0.72 M/s

Questions:

1. What is the reaction order for Cl2 (the value of y)?

2. What is the reaction order for NO (the value of x)?

3. What is the overall rate law for the reaction?

4. Calculate the value of the rate constant (k).

Practice Problem 2: The Zero Order Case

Consider the reaction: A + B -> Products

Trial 1: [A] = 1.0 M, [B] = 1.0 M, Initial Rate = 0.05 M/s

Trial 2: [A] = 2.0 M, [B] = 1.0 M, Initial Rate = 0.05 M/s

Trial 3: [A] = 1.0 M, [B] = 2.0 M, Initial Rate = 0.10 M/s

Questions:

1. What is the reaction order for A (the value of x)?

2. What is the reaction order for B (the value of y)?

3. Write the rate law for this reaction.

Answer Key

Problem 1 Answers:

1. Order for Cl2 (y): Compare Trials 1 and 2. [NO] stays the same, but [Cl2] doubles. The rate also doubles (0.18 to 0.36). This means y = 1.

2. Order for NO (x): Compare Trials 1 and 3. [Cl2] stays the same, but [NO] doubles. The rate quadruples (0.18 to 0.72). Since 2 squared is 4, x = 2.

3. Rate Law: Rate = k[NO]^2[Cl2]^1

4. Rate Constant (k): Use Trial 1 data. 0.18 = k[0.10]^2[0.10]. Solving for k gives 180 M^-2s^-1.

Problem 2 Answers:

1. Order for A (x): Compare Trials 1 and 2. [A] doubles, but the rate stays exactly the same. This means x = 0.

2. Order for B (y): Compare Trials 1 and 3. [B] doubles and the rate doubles. This means y = 1.

3. Rate Law: Rate = k[B] (Reactant A is not included because its order is 0). 

Chemistry - Kinetics

Chemical kinetics is the area of chemistry that deals with the speeds, or rates, of chemical reactions. While thermodynamics tells us if a reaction can happen, kinetics tells us how fast it will happen and the specific steps the molecules take to get there.

The Collision Theory
The foundation of kinetics is the collision theory. For a reaction to occur, reactant particles must collide with one another. However, not every collision results in a reaction. To be successful, a collision must meet two criteria:

Collision Orientation: The molecules must hit each other in the correct position so that the right atoms can bond.

Activation Energy: The particles must collide with enough kinetic energy to break existing chemical bonds.

The activation energy is the minimum amount of energy required to start a chemical reaction. You can think of it like a hill that molecules must climb over before they can roll down the other side into the product state.

Factors Affecting Reaction Rates
Several factors can change how frequently or how forcefully particles collide, which in turn changes the reaction rate:

Concentration: Increasing the concentration of reactants means there are more particles in the same amount of space. This leads to more frequent collisions. In gases, increasing the pressure has the same effect.

Temperature: Increasing the temperature makes particles move faster. This increases the frequency of collisions and, more importantly, ensures that a higher percentage of those collisions have enough energy to surpass the activation energy.

Surface Area: For reactions involving solids, breaking the solid into smaller pieces increases the surface area. This exposes more particles to the other reactants, leading to more frequent collisions.

Catalysts: A catalyst is a substance that speeds up a reaction without being consumed. It works by providing an alternative pathway for the reaction that has a lower activation energy. Because the "hill" is lower, more particles have enough energy to cross it.

Reaction Mechanisms and the Rate-Determining Step
Most chemical reactions do not happen in one single step. Instead, they occur through a series of simpler steps called a reaction mechanism. Each individual step is called an elementary step.

Even if most steps in a mechanism are fast, the overall speed of the reaction is limited by the slowest step. This is known as the rate-determining step. It is like a funnel; no matter how wide the top is, the water can only flow as fast as the narrow neck allows.

Rate Laws
Chemists express the relationship between the rate of a reaction and the concentration of reactants using a mathematical equation called a rate law. A typical rate law looks like this: Rate = k[A]^x [B]^y.

In this equation, k is the rate constant, which is specific to each reaction at a certain temperature. The brackets [A] and [B] represent the molar concentrations of the reactants. The exponents (x and y) are called the reaction orders and must be determined by experiments, not just by looking at the balanced equation. 

Chemistry - Entropy and Gibbs Free Energy

To understand why some chemical reactions happen spontaneously (like a nail rusting) while others don't (like a pile of rust turning back into a nail), we look at two main factors: Entropy and Gibbs Free Energy.

1. Entropy (S): The Measure of Disorder

Entropy is often described as a measure of randomness or disorder in a system. In chemistry, it is more accurately described as the number of ways energy can be distributed among particles.

Low Entropy: Particles are organized, confined, or in a fixed position, like a solid crystal.

High Entropy: Particles are spread out, moving rapidly, or disorganized, like a gas.

The Second Law of Thermodynamics states that the total entropy of the universe is always increasing. This means nature naturally tends toward a state of higher disorder.

Factors that Increase Entropy:

Phase Changes: Moving from solid to liquid to gas.

Temperature: Increasing temperature makes particles move faster and more randomly.

Dissolving: Breaking a solid solute into ions in a solution.

Number of Moles: If a reaction produces more moles of gas than it started with, entropy increases.

2. Gibbs Free Energy (G): The Deciding Factor

While entropy tells us about disorder, it doesn't tell us the whole story. We also have to consider Enthalpy (H), which is the heat energy of the system.

Nature generally prefers low energy (exothermic reactions) and high disorder (increased entropy). Gibbs Free Energy combines these two ideas into one value to determine if a reaction is spontaneous, meaning it can occur without a continuous input of energy.

The Gibbs Equation:

Change in Gibbs Free Energy = (Change in Enthalpy) - (Temperature in Kelvin * Change in Entropy)

Change in G: Change in Free Energy (kJ/mol)

Change in H: Change in Enthalpy (Heat)

T: Temperature (in Kelvin)

Change in S: Change in Entropy

3. Is the Reaction Spontaneous?

The sign of the change in Gibbs Free Energy tells you if the reaction will "go" under specific conditions:

If the change in G is negative (less than 0): Energy is released and available to do work. The reaction is spontaneous.

If the change in G is positive (greater than 0): Energy must be added for the reaction to occur. The reaction is non-spontaneous.

If the change in G equals 0: The system has reached a state of balance called equilibrium.

The Tug-of-War:

Sometimes enthalpy and entropy disagree. For example, melting ice is endothermic (it absorbs heat, which nature usually resists), but it results in higher entropy because liquid water is more disordered than ice. At room temperature, the entropy gain is large enough to make the change in Gibbs Free Energy negative, so ice melts spontaneously. 

Chemistry - Calculating Standard Enthalpy Changes Using Standard Heats of Formation

Reference Table: Standard Heats of Formation (in kJ/mol)

C3H8(g): -103.8

CO2(g): -393.5

H2O(l): -285.8

H2O(g): -241.8

CH4(g): -74.8

NH3(g): -46.1

NO(g): +90.2

Fe2O3(s): -824.2

Al2O3(s): -1675.7

CaO(s): -635.1

CaCO3(s): -1206.9

C2H5OH(l): -277.7

HCl(g): -92.3

Note: Elements in their standard state (such as O2, Fe, and H2) have a heat of formation of 0 kJ/mol.

Formula: Enthalpy of reaction = (Sum of heats of formation of products) - (Sum of heats of formation of reactants) 

Practice Problems

1. Propane Combustion: Calculate the enthalpy change for the combustion of 1 mole of propane gas:

   C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)

2. Methane Combustion: Calculate the enthalpy change for the combustion of methane:

   CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)

3. Thermite Reaction: Determine the heat released when aluminum reacts with iron(III) oxide:

   2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s)

4. Ammonia Oxidation: Calculate the enthalpy change for the following reaction:

   4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

5. Decomposition of Limestone: How much heat is required (or released) during the decomposition of calcium carbonate?

   CaCO3(s) -> CaO(s) + CO2(g)

6. Ethanol Combustion: Calculate the enthalpy change for the combustion of liquid ethanol:

   C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)

7. Synthesis of Hydrogen Chloride: Calculate the enthalpy change for:

   H2(g) + Cl2(g) -> 2HCl(g)

8. Rusting of Iron: Calculate the enthalpy change for the formation of rust:

   4Fe(s) + 3O2(g) -> 2Fe2O3(s)

9. Formation of Water: Using the table, what is the enthalpy change for the reaction of hydrogen and oxygen gas to produce liquid water?

   2H2(g) + O2(g) -> 2H2O(l)

10. Reverse Reaction: If the enthalpy change for a reaction (A + B -> C) is -150 kJ, what is the enthalpy change for the reverse reaction (C -> A + B)?

Answer Key

1. -2219.9 kJ

2. -802.3 kJ

3. -851.5 kJ

4. -905.2 kJ

5. +178.3 kJ

6. -1366.7 kJ

7. -184.6 kJ

8. -1648.4 kJ

9. -571.6 kJ

10. +150 kJ 

Chemistry - The First Law of Thermodynamics and the Heat Equation

Energy is the universe’s most stubborn currency—it can be moved, traded, or changed into a different form, but it can never be created out of thin air or truly destroyed. For a high school chemist, understanding how energy flows in a system is the difference between a successful experiment and an unexpected boom.

The First Law of Thermodynamics is essentially the Law of Conservation of Energy applied to heat and work. It tells us that the change in the internal energy of a system (delta U) is the sum of the heat (q) added to the system and the work (w) done on or by the system. The formula is: delta U = q + w.

Heat (q) is thermal energy transferred between objects due to a temperature difference. Work (w) is energy used to move something, like a piston in an engine. In most chemistry labs, we focus primarily on the heat component. If a reaction releases heat, it is exothermic; if it absorbs heat, it is endothermic.

To calculate exactly how much energy is being transferred as heat during a temperature change, we use the Specific Heat Equation: q = (m) (c) (delta T).

In this equation, q is the amount of heat measured in Joules (J). The variable m is the mass of the substance measured in grams (g). The variable c is the specific heat capacity, measured in J/gC. This is a unique thermal personality for every substance—water has a very high specific heat, while metals have very low ones. Finally, delta T is the change in temperature, calculated as the final temperature minus the initial temperature. Equation: delta T = final temp - initial temp

Practice Problems

1. Heating Water: How much heat is required to raise the temperature of 250 g of water from 20 C to 80 C? (Specific heat of water is 4.184 J/gC).

2. Identifying a Metal: A 50 g sample of an unknown metal releases 1,200 J of heat, causing its temperature to drop by 60 C. What is the specific heat capacity of the metal?

3. The Cooling Process: If 500 J of heat is removed from 25 g of aluminum (c = 0.90 J/gC) at 100 C, what is the final temperature?

4. Mass Matters: A sample of copper (c = 0.385 J/gC) absorbs 2,500 J of heat and its temperature increases by 15 C. What is the mass of the copper?

5. First Law Concept: A system performs 150 J of work on its surroundings and absorbs 400 J of heat. Calculate the change in internal energy (delta U).

Answer Key

1. 62,760 J

2. 0.40 J/gC

3. 77.8 C

4. 432.9 g

5. +250 J

Chemistry - The Ideal Gas Law and Stoichiometry

Understanding the Relationship

Gas stoichiometry combines the principles of chemical reactions with the properties of gases. In high school chemistry, you have already learned how to use the Ideal Gas Law (PV = nRT) to find properties of a single gas. Stoichiometry allows you to use the balanced chemical equation to relate the amount of one substance to another. When these two concepts meet, you can calculate the volume of a gas produced in a reaction or determine how much of a solid reactant is needed to produce a specific pressure of gas.

The key link between these two worlds is the mole (n). Stoichiometry uses mole ratios from balanced equations, and the Ideal Gas Law solves for moles based on pressure (P), volume (V), and temperature (T).

The Ideal Gas Law Formula: PV = nRT

P is Pressure (commonly in atm or kPa)

V is Volume (must be in Liters)

n is Number of Moles

R is the Ideal Gas Constant (0.0821 L atm / mol K or 8.314 L kPa / mol K)

T is Temperature (must be in Kelvin; K = Celsius + 273)

Steps for Gas Stoichiometry Problems

1. Write or check the balanced chemical equation.

2. Identify the given information. Is it gas data (P, V, T) or mass/mole data?

3. Convert your "Given" to Moles. If you start with gas data, use n = PV / RT. If you start with mass, use the molar mass.

4. Use the Mole Ratio from the balanced equation to find the moles of the "Unknown" substance. 

5. Convert the Moles of the Unknown to the final units requested (Volume, Mass, or Pressure).

Practice Problems

1. The combustion of propane follows the equation: C3H8 + 5 O2 --> 3 CO2 + 4 H2O. If 5.00 grams of propane (C3H8) is burned, what volume of CO2 gas will be produced at 1.00 atm and 298 K?

2. Solid calcium carbonate decomposes into calcium oxide and carbon dioxide gas: CaCO3 --> CaO + CO2.  How many grams of CaCO3 are needed to produce 15.0 L of CO2 at a pressure of 105 kPa and a temperature of 300 K?

3. Hydrogen gas is produced by reacting zinc with sulfuric acid: Zn + H2SO4 --> ZnSO4 + H2. If 12.0 grams of zinc react completely, what will be the pressure of the hydrogen gas if it is collected in a 4.00 L flask at 27 degrees Celsius?

4. In the reaction 2 NaN3 --> 2 Na + 3 N2, used in car airbags, how many grams of NaN3 are required to provide 65.0 L of nitrogen gas at 1.15 atm and 30 degrees Celsius?

5. Ammonia is produced via the Haber process: N2 + 3 H2 --> 2 NH3. If 20.0 L of nitrogen gas at 200 kPa and 500 K reacts with excess hydrogen, how many moles of ammonia are produced?

Answer Key

1. 8.32 L of CO2

   Steps: Convert 5.00g C3H8 to moles (0.113 mol). Use 3:1 ratio to get 0.339 mol CO2. Use V = nRT/P.

2. 63.2 grams of CaCO3

   Steps: Use n = PV/RT to find moles of CO2 (0.631 mol). Use 1:1 ratio to find moles of CaCO3. Multiply by molar mass (100.09 g/mol).

3. 1.13 atm

   Steps: Convert 12.0g Zn to moles (0.184 mol). Use 1:1 ratio to get 0.184 mol H2. Use P = nRT/V with T=300K.

4. 130.5 grams of NaN3

   Steps: Use n = PV/RT to find moles of N2 (3.01 mol). Use 2:3 ratio to find moles of NaN3 (2.007 mol). Multiply by molar mass (65.01 g/mol).

5. 1.93 moles of NH3

   Steps: Use n = PV/RT to find moles of N2 (0.963 mol). Use 2:1 mole ratio to find moles of NH3.