3.2 Chemical Equilibrium, Acids/Bases, and Ideal Gas Laws

3.2 Chemical Equilibrium, Acids/Bases, and Ideal Gas Laws

Chemical systems in nature and industrial engineering reach state balances governed by thermodynamics and kinetics. Understanding chemical equilibrium, aqueous acid-base buffers, and fluid gas behavior under varying pressure and temperature is vital for FE exam success.

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1. Chemical Equilibrium Constants and Le Chatelier's Principle

Many chemical reactions do not proceed to 100% completion; instead, they reach a state of dynamic chemical equilibrium, where the forward reaction rate equals the reverse reaction rate, leaving net reactant and product concentrations constant over time.

The Equilibrium Constant ($K_c$ and $K_p$)

For a general reversible gas-phase or aqueous reaction: $a A + b B \rightleftharpoons c C + d D$

NCEES Formula: Concentration Equilibrium Constant ($K_c$)

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ where $[i]$ denotes molar concentration ($\text{mol/L}$). Pure solids ($s$) and pure liquid solvents ($l$) have an activity of 1.0 and are omitted from the equilibrium expression.

NCEES Formula: Pressure Equilibrium Constant ($K_p$)

For gas-phase reactions expressed in partial pressures (atm or bar): $K_p = \frac{P_C^c P_D^d}{P_A^a P_B^b}$

NCEES Formula: $K_p$ to $K_c$ Relationship

$K_p = K_c (R T)^{\Delta n}$ where:

• $R$ = ideal gas constant ($0.08206 \text{ L}\cdot\text{atm/(mol}\cdot\text{K)}$ or $8.314 \text{ J/(mol}\cdot\text{K)}$)
• $T$ = absolute temperature in Kelvin ($\text{K} = ^\circ\text{C} + 273.15$)
• $\Delta n = (c + d) - (a + b)$ = change in gaseous moles (moles gas products - moles gas reactants)

Reaction Quotient ($Q$) and Reaction Direction

Evaluating the expression using non-equilibrium initial concentrations yields the reaction quotient ($Q$):

• $Q < K$: Reaction shifts right (toward products) to reach equilibrium.
• $Q = K$: System is at dynamic equilibrium.
• $Q > K$: Reaction shifts left (toward reactants) to reach equilibrium.

Le Chatelier's Principle

If an external stress is applied to a system at dynamic equilibrium, the system adjusts itself to partially offset that stress:

1. Concentration: Adding a reactant or removing a product shifts the equilibrium to the right.
2. Pressure / Volume: Decreasing volume (increasing pressure) shifts equilibrium toward the side with fewer moles of gas.
3. Temperature:
   • For an exothermic reaction ($\Delta H < 0$, heat is a product): Increasing temperature shifts equilibrium left (decreases $K$).
   • For an endothermic reaction ($\Delta H > 0$, heat is a reactant): Increasing temperature shifts equilibrium right (increases $K$).

2. Acid-Base Equilibrium, pH, and Buffers

Aqueous chemical processing relies heavily on acid-base equilibria. Under the Brønsted-Lowry definition, an acid is a proton ($H^+$) donor, and a base is a proton ($H^+$) acceptor.

Water Autoionization and the pH Scale

Water undergoes autoionization: $\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$. At $25^\circ\text{C}$ ($298.15\text{ K}$), the ion-product constant of water ($K_w$) is:

NCEES Formula: Water Ion Product

$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

NCEES Formula: pH and pOH Definitions

$pH = -\log_{10}[H^+] \quad \text{and} \quad pOH = -\log_{10}[OH^-]$ $pH + pOH = 14.00$

Weak Acid Dissociation ($K_a$)

Strong acids (e.g., $\text{HCl}, \text{HNO}_3, \text{H}_2\text{SO}_4$) dissociate 100% in water. Weak acids ($HA$) reach partial equilibrium: $HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$ $K_a = \frac{[H^+][A^-]}{[HA]}$ where $K_a$ is the acid dissociation constant. Larger $K_a$ values indicate stronger weak acids.

Buffer Solutions and Henderson-Hasselbalch Equation

A buffer solution consists of a weak acid ($HA$) and its conjugate base ($A^-$) in comparable concentrations. Buffers resist significant pH changes upon addition of small amounts of strong acid or base.

NCEES Formula: Henderson-Hasselbalch Equation

$pH = pK_a + \log_{10} \left( \frac{[A^-]}{[HA]} \right)$ where $pK_a = -\log_{10}(K_a)$.

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Worked Engineering Example: Weak Acid and Buffer pH Calculations

Problem:

1. Calculate the pH of a $0.150\text{ M}$ solution of acetic acid ($\text{CH}_3\text{COOH}$, $K_a = 1.80 \times 10^{-5}$).
2. Calculate the pH after adding sodium acetate ($\text{CH}_3\text{COONa}$) to the solution such that $[A^-] = 0.250\text{ M}$.

Solution:

Part 1: Weak Acid Equilibrium (ICE Table): Reaction: $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$

• Initial: $[HA] = 0.150$, $[H^+] = 0$, $[A^-] = 0$
• Change: $-x$, $+x$, $+x$
• Equilibrium: $0.150 - x$, $x$, $x$

$K_a = \frac{x^2}{0.150 - x} = 1.80 \times 10^{-5}$ Assuming $x \ll 0.150$: $x^2 = 1.80 \times 10^{-5} \times 0.150 = 2.70 \times 10^{-6} \implies x = 1.643 \times 10^{-3}\text{ M}$ Check 5% rule: $(1.643 \times 10^{-3} / 0.150) \times 100% = 1.1% < 5%$ (valid assumption). $pH = -\log_{10}(1.643 \times 10^{-3}) = 2.784 \approx 2.78$

Part 2: Buffer pH via Henderson-Hasselbalch Equation: $pK_a = -\log_{10}(1.80 \times 10^{-5}) = 4.745$ $pH = pK_a + \log_{10} \left( \frac{[A^-]}{[HA]} \right) = 4.745 + \log_{10} \left( \frac{0.250}{0.150} \right)$ $pH = 4.745 + \log_{10}(1.6667) = 4.745 + 0.222 = 4.967 \approx 4.97$ Conclusion: Pure $0.150\text{ M}$ acetic acid has a pH of $2.78$, whereas the buffered conjugate system raises and stabilizes the pH at $4.97$.

3. Ideal Gas Laws, Partial Pressures, and Real Gas Behavior

Gas thermodynamics forms a core topic on the FE exam, connecting fluid behavior with chemical process design.

The Ideal Gas Law

Ideal gases assume zero molecular volume and zero intermolecular attractive forces.

NCEES Formula: Ideal Gas Law

$P V = n R T$

where:

• $P$ = absolute pressure ($\text{Pa, atm, kPa}$)
• $V$ = volume ($\text{m}^3, \text{L}$)
• $n$ = moles of gas ($\text{mol, kmol}$)
• $R$ = universal gas constant ($8.314 \text{ J/(mol}\cdot\text{K)} = 0.08206 \text{ L}\cdot\text{atm/(mol}\cdot\text{K)}$)
• $T$ = absolute temperature ($\text{K}$)

Gas Density Formula

$\rho = \frac{m}{V} = \frac{P \cdot M}{R \cdot T}$ where $M$ is molar mass.

Dalton's Law of Partial Pressures

In a mixture of non-reacting ideal gases, total pressure equals the sum of partial pressures exerted by individual gas species: $P_{\text{total}} = \sum P_i = P_1 + P_2 + \dots + P_k$ $P_i = \chi_i P_{\text{total}}$ where $\chi_i = n_i / n_{\text{total}}$ is the mole fraction of gas $i$.

Non-Ideal (Real) Gas Behavior: Van der Waals Equation

At high pressures ($P > 10\text{ atm}$) or low temperatures ($T \approx T_{critical}$), real gas molecules exhibit non-zero physical volume and attractive intermolecular forces, causing deviation from ideal behavior.

NCEES Formula: Van der Waals Equation

$\left( P + \frac{a n^2}{V^2} \right) (V - n b) = n R T$

where:

• $a$ = empirical constant accounting for intermolecular attractive forces (reduces observed wall impact pressure).
• $b$ = empirical constant accounting for the finite physical volume occupied by gas molecules (reduces free volume).

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Worked Engineering Example: Ideal vs. Real Gas Pressure Comparison

Problem: Determine the pressure exerted by $2.00\text{ moles}$ of carbon dioxide gas ($\text{CO}_2$) confined within a $5.00\text{ L}$ vessel at $300.0\text{ K}$ using:

1. The Ideal Gas Law
2. The Van der Waals equation ($a = 3.59 \text{ L}^2\cdot\text{atm/mol}^2$, $b = 0.0427 \text{ L/mol}$)

Solution:

Part 1: Ideal Gas Law: $P_{\text{ideal}} = \frac{n R T}{V} = \frac{(2.00\text{ mol})(0.08206 \text{ L}\cdot\text{atm/(mol}\cdot\text{K)})(300.0\text{ K})}{5.00\text{ L}} = 9.847\text{ atm}$

Part 2: Van der Waals Equation: Rearranging for pressure $P$: $P = \frac{n R T}{V - n b} - \frac{a n^2}{V^2}$ Compute corrected volume term ($V - nb$): $V - n b = 5.00\text{ L} - (2.00\text{ mol})(0.0427\text{ L/mol}) = 5.00 - 0.0854 = 4.9146\text{ L}$ Compute kinetic pressure component: $P_{\text{kinetic}} = \frac{(2.00)(0.08206)(300.0)}{4.9146} = \frac{49.236}{4.9146} = 10.0183\text{ atm}$ Compute attractive correction term ($a n^2 / V^2$): $P_{\text{attraction}} = \frac{(3.59)(2.00)^2}{(5.00)^2} = \frac{14.36}{25.00} = 0.5744\text{ atm}$ Net real pressure: $P_{\text{real}} = 10.0183 - 0.5744 = 9.4439\text{ atm} \approx 9.44\text{ atm}$ Conclusion: Intermolecular attractions reduce real $\text{CO}_2$ pressure to $9.44\text{ atm}$, representing a $4.1%$ downward deviation from ideal gas behavior.

4. Summary Table of Gas Laws & Acid-Base Relationships