Stoichiometry, Gas Laws, and Kinetic-Molecular Theory: Study Notes

Spontaneity, Enthalpy, Entropy, and Gibbs Free Energy

Spontaneous Processes

A spontaneous process is one that, once started, proceeds on its own without continuous external influence.

• Enthalpy (ΔH): Decreases in enthalpy (exothermic processes) favor spontaneity, but do not guarantee it.

• Entropy (S): A measure of system "randomness" or "disorder." Increases in entropy favor spontaneity, but do not guarantee it.

• Temperature: Also affects spontaneity.

• Gibbs Free Energy (ΔG): Spontaneity is determined via the Gibbs free-energy change:

$\Delta G = \Delta H - T\Delta S$

• If $\Delta G < 0$, the process is spontaneous.

Gas Laws and Stoichiometry

The Ideal Gas Law

The ideal gas law combines all the information of the individual gas laws and is used to calculate gas pressure, volume, temperature, and number of moles:

$PV = nRT$

• P: Pressure (atm)

• V: Volume (L)

• n: Number of moles

• R: Universal gas constant ($0.08206$ L·atm/mol·K)

• T: Temperature (K)

When stoichiometry is combined with the gas laws, the ideal gas law is used to convert between moles, pressure, volume, and temperature of the gas.

Limiting Reactants

In chemical reactions, the limiting reactant is the reactant present in a lesser stoichiometric amount, which limits the amount of product that forms.

• Identify the limiting reactant by calculating the amount of product each reactant could produce.

• The reactant that produces the least amount of product is the limiting reactant.

Limiting Reactant Example

Given: 35.6 L of NH3 and 40.5 L of O2 at STP react to form NO and H2O.

Balanced equation: $4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(l)$

1. Convert volume of each reactant to moles using the ideal gas law.

2. Use stoichiometry to determine grams of NO produced from each reactant.

3. The reactant producing the least NO is limiting; the amount of NO produced is the answer.

Example Calculations:

$n_{NH_3} = \frac{PV}{RT} = \frac{(1\,atm)(35.6\,L)}{(0.08206\,L\cdot atm/mol\cdot K)(273.15\,K)} = 1.5882\,mol\,NH_3$

$n_{O_2} = \frac{PV}{RT} = \frac{(1\,atm)(40.5\,L)}{(0.08206\,L\cdot atm/mol\cdot K)(273.15\,K)} = 1.8069\,mol\,O_2$

Calculate grams NO from each reactant:

$x\,g\,NO = n_{NH_3} \times \frac{4\,mol\,NO}{4\,mol\,NH_3} \times 30.0061\,g/mol = 47.7\,g\,NO$

$x\,g\,NO = n_{O_2} \times \frac{4\,mol\,NO}{5\,mol\,O_2} \times 30.0061\,g/mol = 43.4\,g\,NO$

O2 is the limiting reactant; the reaction produces 43.4 g NO.

Stoichiometry and Gas Law Practice Problems

• Use $PV = nRT$ to solve for unknowns in reactions involving gases.

• Example: Calculate the volume of H2 produced from Zn and HCl at given conditions.

Dalton’s Law of Partial Pressures

Definition and Application

Dalton’s law of partial pressures states that in a mixture of unreacting gases, the total pressure is the sum of the partial pressures of the individual gases:

$P_{total} = P_1 + P_2 + P_3 + \ldots$

• Each gas in a mixture behaves as if it were the only gas present.

• Each gas has the same temperature and occupies the same volume.

• The total pressure can be rewritten as:

$P_{total} = (n_1 + n_2 + \ldots) \frac{RT}{V} = n_{total} \frac{RT}{V}$

Mole Fraction

The mole fraction ($X_i$) is the number of moles of a component divided by the total number of moles in the mixture:

$X_i = \frac{n_i}{n_{total}}$

Partial pressure of a component:

$P_i = X_i P_{total}$

Partial Pressure Example

Given a mixture of CO and SO2 with total pressure, calculate the partial pressure of CO:

• Find moles of each gas.

• Calculate mole fraction of CO.

• Multiply by total pressure to get partial pressure.

Example: $P_{CO} = X_{CO} P_{total} = (0.61)(0.33\,atm) = 0.20\,atm$

Kinetic-Molecular Theory

Postulates

• The volume occupied by individual gas particles is negligible compared to the container volume.

• Gas particles are in constant, random, straight-line motion, except during collisions.

• There are neither attractive nor repulsive forces between particles (ideal gas assumption).

• Energy is exchanged during collisions, but total kinetic energy is constant (elastic collisions).

Distribution of Speeds

Particle speed is best described as a distribution, with the most probable speed directly proportional to absolute temperature.

Graphical representation: As temperature increases, the distribution flattens and shifts to higher speeds.

Molecular-Level Nature of Pressure

• Pressure is due to gas particle collisions with the container walls.

Gas Laws Explained by Kinetic-Molecular Theory

• Boyle’s Law ($V \propto 1/P$): As volume decreases, pressure increases due to more frequent collisions.

• Charles’s Law ($V \propto T$): As temperature increases, molecular speed and collision frequency increase, raising pressure and volume.

• Avogadro’s Law ($V \propto n$): More moles mean more collisions, increasing pressure and volume.

• Amonton’s Law ($P \propto T$): Higher temperature increases collision frequency and energy, raising pressure.

• Dalton’s Law: Total pressure is the sum of partial pressures, each due to the number of particles, not their identity.

Kinetic Energy, Temperature, and Mass

Relationship Between Kinetic Energy and Temperature

• Average kinetic energy of gas particles depends only on temperature:

$E_{kinetic} = \frac{3RT}{2N_A}$

• Kinetic energy also depends on mass and speed:

$E_{kinetic} = \frac{1}{2}mu^2$

Equating the two:

$\frac{3RT}{2N_A} = \frac{1}{2}mu^2$

Conclusion: At the same temperature, heavy particles move more slowly than light particles.

Average Gas Particle Speed

Average speed ($u$) of a gas particle:

$u = \sqrt{\frac{3RT}{m_{molar}}}$

Example: For N2 (28.0134 g/mol) at 298 K:

$u = \sqrt{\frac{3 \times 8.314 \times 298}{0.0280134}}$

Table of average speeds for common gases:

Gas	Molar Mass (g/mol)	Average Speed (m/s)
H2	2.0	1960
He	4.0	1360
H2O	18.0	650
N2	28.0	490
O2	32.0	490
CO2	44.0	415

Diffusion and Effusion of Gases

Definitions

• Diffusion: Mixing of gas molecules due to random motion.

• Effusion: Passage of gas through a small opening into a vacuum.

Graham’s Law of Effusion

The rate of effusion of a gas is inversely proportional to the square root of its molar mass:

$\text{Rate of effusion} \propto \frac{1}{\sqrt{m}}$

Comparing two gases:

$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{m_2}{m_1}}$

Example: The effusion rate of an unknown gas is 1.19 times greater than Ar (39.95 g/mol). Find the molar mass:

$1.19 = \sqrt{\frac{39.95}{m_{unknown}}}$ Solve for $m_{unknown}$.

Real Gases and Deviations from Ideality

Non-Ideal Gas Behavior

• At high pressures, the volume occupied by gas particles is not negligible.

• At high pressures, particle attractions and repulsions become significant.

• The van der Waals equation accounts for these deviations:

$(P + a\frac{n^2}{V^2})(V - nb) = nRT$

• a: Corrects for intermolecular attractions.

• b: Corrects for finite volume of gas particles.

Summary Table: Key Equations and Concepts

Concept	Equation	Notes
Ideal Gas Law	$PV = nRT$	Relates P, V, n, T for ideal gases
Dalton's Law	$P_{total} = P_1 + P_2 + \ldots$	Total pressure is sum of partial pressures
Mole Fraction	$X_i = \frac{n_i}{n_{total}}$	Fractional composition of mixture
Partial Pressure	$P_i = X_i P_{total}$	Pressure due to one component
Kinetic Energy	$E_{kinetic} = \frac{3RT}{2N_A}$	Average kinetic energy per particle
Average Speed	$u = \sqrt{\frac{3RT}{m_{molar}}}$	Root-mean-square speed
Graham's Law	$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{m_2}{m_1}}$	Relative effusion rates
van der Waals	$(P + a\frac{n^2}{V^2})(V - nb) = nRT$	Corrects for real gas behavior

Additional info:

• Some slides included practice questions and worked examples to reinforce concepts.

• Distribution of speeds and molecular-level explanations were supported by diagrams and tables (recreated above).