Chem 1202 - Chapter 15: Chemical Equilibrium

Chem 1202 - General Chemistry 2

Chapter 15 - Chemical Equilibrium

Chapter 15 Learning Objectives

• Explain chemical equilibrium and its relation to reaction rates (Section 15.1).
• Write equilibrium-constant expressions for any reaction (Section 15.2).
• Convert $K<em>c$ to $K</em>p$ and vice versa (Section 15.2).
• Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products in an equilibrium mixture (Section 15.3).
• Manipulate the equilibrium constant to reflect changes in the chemical equation (Section 15.3).
• Write equilibrium-constant expressions for heterogeneous reactions (Section 15.4).
• Calculate equilibrium constants from concentration measurements (Section 15.5).
• Predict the direction of a reaction given the equilibrium constant and concentrations (Section 15.6).
• Calculate equilibrium concentrations (Section 15.6).
• Use Le Châtelier’s principle (Section 15.7).

Section 15.1 - The Concept of Equilibrium

• Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
• As a system approaches equilibrium, both forward and reverse reactions happen.
• At equilibrium, the forward and reverse reactions proceed at the same rate.
• Once equilibrium is achieved, the amount of each reactant and product remains constant.
• Products are being made, but reactants are being made at the same rate.
• Total amount of reactant and product doesn’t change.
• Neither reactant nor product can escape the system.
• There is a ratio of concentration terms that is constant.
• Equation for an Equilibrium Reaction: $N<em>2O</em>4(g) \rightleftharpoons 2NO_2(g)$
• For the forward reaction $N<em>2O</em>4(g) \rightarrow 2NO<em>2(g)$, the rate law is $Rate = k</em>f[N<em>2O</em>4]$.
• For the reverse reaction $2NO<em>2(g) \rightarrow N</em>2O<em>4(g)$, the rate law is $Rate = k</em>r[NO_2]^2$.
• Meaning of Equilibrium:
  • At equilibrium, $Rate<em>f = Rate</em>r$.
  • $k<em>f[N</em>2O<em>4] = k</em>r[NO_2]^2$.
  • $\frac{[NO<em>2]^2}{[N</em>2O<em>4]} = \frac{k</em>f}{k<em>r} = a constant = K</em>c$
• General expression:
  • $\frac{[products]}{[reactants]} = K_c$

Section 15.2 - The Equilibrium Constant

• Haber Process: $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$
  • The equilibrium constant depends on stoichiometry: $K<em>c = \frac{[NH</em>3]^2}{[N<em>2][H</em>2]^3}$.
• It doesn’t matter whether you start with reactants or products; an equilibrium will be reached. $\text{N}<em>2 + 3 \text{H}</em>2 \rightleftharpoons 2 \text{NH}_3$
• Generalized reaction: $aA + bB \rightleftharpoons dD + eE$
  • The equilibrium expression for this reaction would be $K_c = \frac{[D]^d[E]^e}{[A]^a[B]^b} \leftarrow products \leftarrow reactants$
  • Also, since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written as $K<em>p = \frac{P</em>D^dP<em>E^e}{P</em>A^aP_B^b}$.
• Law of Mass Action:
  • $aA + bB \rightleftharpoons dD + eE$
  • $K_c = \frac{[D]^d[E]^e}{[A]^a[B]^b} = \frac{products}{reactants}$
  • $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$
  • $K<em>c = \frac{[NH</em>3]^2}{[N<em>2]^1[H</em>2]^3} = \frac{products}{reactants}$
• Equilibrium Constant Demonstrated:
  • The ratio of $[NO<em>2]^2$ to $[N</em>2O_4]$ remains constant at a given temperature regardless of initial concentrations.
  • $N<em>2O</em>4 \rightleftharpoons 2NO_2$
• More with Gases and Equilibrium:
  • For gases, $PV = nRT$ (the ideal-gas law).
  • Rearranging, $P = (\frac{n}{V})RT$; $\frac{n}{V}$ is $\text{[ ]}$.
  • The result is $K<em>p = K</em>c(RT)^{\Delta n}$.
  • where $\Delta n = \text{moles of gaseous product } - \text{ moles of gaseous reactant}$

Section 15.3 - Understanding and Working with Equilibrium Constants

• Magnitude of K:
  • If K >> 1, the reaction favors products; products predominate at equilibrium.
  • If K << 1, the reaction favors reactants; reactants predominate at equilibrium.
• Magnitude of $K_{eq}$ should immediately tell you in what direction a reaction is going to proceed and how far it will go before reaching equilibrium.
  • K_c >> 1 reaction will go mainly to products
  • $K_c \sim 1$ reaction will produce roughly equal amounts of product and reactant
  • K_c << 1 reaction will go mainly to reactants
• Exercises:
  • $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH<em>3(g) \quad K</em>c = 0.001$ (Reactants)
  • $2SO<em>3(g) \rightleftharpoons 2SO</em>2(g) + O<em>2(g) \quad K</em>c = 2$ (Products)
  • $2HBr(g) + Cl<em>2(g) \rightleftharpoons 2HCl(g) + Br</em>2(g) \quad K_c = 10^4$ (Products)
  • $2H<em>2O(g) \rightleftharpoons 2H</em>2(g) + O<em>2(g) \quad K</em>c = 10^{-28}$ (Reactants)
  • $N<em>2O</em>4(g) \rightleftharpoons 2NO<em>2(g) \quad K</em>c = 1$ (Roughly equal)
  • $H<em>2(g) + Cl</em>2(g) \rightleftharpoons 2HCl(g) \quad K_c = 10^{44}$ (Products)
  • $PCl<em>3(soln) + Cl</em>2(soln) \rightleftharpoons PCl<em>5(soln) \quad K</em>c = 0.1$ (Reactants)
  • $2HCl(g) + Br<em>2(g) \rightleftharpoons 2HBr(g) + Cl</em>2(g) \quad K_c = 10^{-4}$ (Reactants)
• The Direction of the Chemical Equation and K
  • The equilibrium constant of a reverse reaction is the reciprocal of the forward reaction's equilibrium constant.
  • $N<em>2O</em>4(g) \rightleftharpoons 2NO<em>2(g) \quad K</em>c = \frac{[NO<em>2]^2}{[N</em>2O_4]} = 0.212 \text{ at } 100 \degree C$
  • $2NO<em>2(g) \rightleftharpoons N</em>2O<em>4(g) \quad K</em>c = \frac{[N<em>2O</em>4]}{[NO_2]^2} = 4.72 \text{ at } 100 \degree C$
• Stoichiometry and Equilibrium Constants
  • To find the new equilibrium constant when the equation is multiplied by a number, raise the original equilibrium constant to that power.
  • $N<em>2O</em>4(g) \rightleftharpoons 2NO<em>2(g) \quad K</em>c = \frac{[NO<em>2]^2}{[N</em>2O_4]} = 0.212 \text{ at } 100 \degree C$
  • $2N<em>2O</em>4(g) \rightleftharpoons 4NO<em>2(g) \quad K</em>c = \frac{[NO<em>2]^4}{[N</em>2O_4]^2} = (0.212)^2 \text{ at } 100 \degree C$
• Consecutive Equilibria
  • When two consecutive equilibria occur, the equations can be added to give a single equilibrium.
  • The equilibrium constant of the new reaction is the product of the two constants: $K<em>3 = K</em>1 \times K_2$
  • Example:
    • $2NOBr \rightleftharpoons 2NO + Br<em>2 \quad K</em>1 = 0.014$
    • $Br<em>2 + Cl</em>2 \rightleftharpoons 2BrCl \quad K_2 = 7.2$
    • $2NOBr + Cl<em>2 \rightleftharpoons 2NO + 2BrCl \quad K</em>3 = K<em>1 \times K</em>2 = 0.014 \times 7.2 = 0.10$

Section 15.4 - Heterogeneous Equilibria

• Homogeneous vs. Heterogeneous
  • Homogeneous equilibria occur when all reactants and products are in the same phase.
  • Heterogeneous equilibria occur when something in the equilibrium is in a different phase.
  • The value used for the concentration of a pure substance is always 1.
• The Decomposition of $CaCO_3$—A Heterogeneous Equilibrium
  • $CaCO<em>3(s) \rightleftharpoons CaO(s) + CO</em>2(g)$
  • $K<em>c = [CO</em>2]$ and $K<em>p = P</em>{CO_2}$
• Examples:
  • $CaO(s) + CO<em>2(g) \rightleftharpoons CaCO</em>3(s) \quad K<em>c = \frac{1}{[CO</em>2]}$
  • $Ag^+(aq) + Cl^-(aq) \rightleftharpoons AgCl(s) \quad K_c = \frac{1}{[Ag^+][Cl^-]}$
  • $Br<em>2(l) + Ni(CO)</em>4(l) \rightleftharpoons NiBr<em>2(s) + 4CO(g) \quad K</em>c = [CO]^4$
  • $H<em>2CO</em>3(aq) \rightleftharpoons H<em>2O(l) + CO</em>2(g) \quad K<em>c = \frac{[CO</em>2]}{[H<em>2CO</em>3]}$
  • $HCl(g) + H<em>2O(l) \rightleftharpoons H^+(aq) + Cl^-(aq) \quad K</em>c = \frac{[H^+][Cl^-]}{[HCl]}$

Section 15.5 - Calculating Equilibrium Constants

• Deducing Equilibrium Concentrations
  1. Tabulate all known initial and equilibrium concentrations.
  2. For anything for which initial and equilibrium concentrations are known, calculate the change.
  3. Use the balanced equation to find change for all other reactants and products.
  4. Use initial concentrations and changes to find equilibrium concentration of all species.
  5. Calculate the equilibrium constant using the equilibrium concentrations.
• Example:
  • A closed system initially containing $1.000 \times 10^{-3} M$ $H<em>2$ and $2.000 \times 10^{-3} M$ $I</em>2$ at $448 \degree C$ is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is $1.87 \times 10^{-3} M$. Calculate $K<em>c$ at $448 \degree C$ for the reaction taking place, which is $H</em>2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
  • $K<em>c = \frac{[HI]^2}{[H</em>2][I_2]} = \frac{(1.87 \times 10^{-3})^2}{(6.5 \times 10^{-5})(1.065 \times 10^{-3})} = 51$

Section 15.6 - Applications of Equilibrium Constants

• Is a Mixture in Equilibrium? Which Way Does the Reaction Go?
  • To answer these questions, we calculate the reaction quotient, Q.
  • Q looks like the equilibrium constant, K, but the values used to calculate it are the current conditions, not necessarily those for equilibrium.
  • To calculate Q, one substitutes the concentrations or pressures of reactants and products at any given time during a reaction into the equilibrium expression.
• Mixture in Equilibrium:
  • Rates change with time, reach equilibrium, and equilibrium: rates are equal
  • $rate<em>{forward} = rate</em>{reverse}$
• Rates are not equal → not in Equilibrium:
  • $aA + bB \rightleftharpoons dD + eE$
  • $rate<em>{forward} \ne rate</em>{reverse}$
  • Replace $K_c$ by reaction quotient Q
  • $Q = \frac{[D]^d[E]^e}{[A]^a[B]^b} = \frac{products}{reactants}$
• $K_{eq}$ and Q look the same:
  • $aA + bB \rightleftharpoons dD + eE$
  • $rate<em>{forward} = rate</em>{reverse} \quad K_c = \frac{[D]^d[E]^e}{[A]^a[B]^b} = \frac{products}{reactants}$
  • $rate<em>{forward} \ne rate</em>{reverse} \quad Q = \frac{[D]^d[E]^e}{[A]^a[B]^b} = \frac{products}{reactants}$
• Why is Q important?
  • By comparing Q to $K_c$, one can tell in which direction a reaction will go to reach a state of equilibrium:
  • $wA + xB \rightleftharpoons yC + zD$
  • $Q = \frac{[C]^y[D]^z}{[A]^w[B]^x} = \frac{products}{reactants}$
  • Q > K_c reverse reaction will be spontaneous
  • $Q = K_c$ reaction @ equilibrium
  • Q < K_c forward reaction will be spontaneous
• Example:
  • If the initial concentrations of all species are 1 M, which way will the following reaction proceed to reach equilibrium?
  • $CO<em>2(g) + H</em>2(g) \rightleftharpoons CO(g) + H<em>2O(g) \quad K</em>c = 4.4$
  • $Q = \frac{[CO][H<em>2O]}{[CO</em>2][H<em>2]} = \frac{1 \times 1}{1 \times 1} = 1 < 4.4 = K</em>c$
• If you are ever given a problem where the product and reactant concentrations are all non-zero, MUST calculate Q and compare it to Keq to figure out which way the reaction has to go to reach equilibrium.
• Comparing Q and K
  • Nature wants Q = K.
  • If Q < K, nature will make the reaction proceed to products.
  • If Q = K, the reaction is in equilibrium.
  • If Q > K, nature will make the reaction proceed to reactants.
• Calculating Equilibrium Concentrations
  • If you know the equilibrium constant, you can find equilibrium concentrations from initial concentrations and changes (based on stoichiometry).
  • You will set up a table to find the equilibrium concentration, but the “change in concentration” row will simply be a factor of “x” based on the stoichiometry.
• Examples:
  • A 1.000 L flask is filled with 1.000 mol of $H<em>2(g)$ and 2.000 mol of $I</em>2(g)$ at $448 \degree C$. Given a $K<em>c$ of 50.5 at $448 \degree C$, what are the equilibrium concentrations of $H</em>2$, $I_2$, and HI?
    • $H<em>2(g) + I</em>2(g) \rightleftharpoons 2HI(g)$
    • $K<em>c = \frac{[HI]^2}{[H</em>2][I_2]} = \frac{(2x)^2}{(1.000 - x)(2.000 - x)} = 50.5 \implies x = 0.935$
    • $[H<em>2]</em>{eq} = 1.000 - 0.935 = 0.065 M$
    • $[I<em>2]</em>{eq} = 2.000 - 0.935 = 1.065 M$
    • $[HI]_{eq} = 2 \times 0.935 = 1.87 M$
      Attention: Always need to provide concentration, Calculation of x not enough

Section 15.7 - Le Châtelier's Principle

• “If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”
• Change in Reactant or Product Concentration
  • Adding a reaction component will result in it being used up.
  • Removing a reaction component will result in it being produced.
• Change in Volume or Pressure
  • When gases are involved in an equilibrium, a change in pressure or volume will affect equilibrium: Higher volume or lower pressure favors the side of the equation with more moles (and vice versa).
• Example: Doubling Pressure (or Halving Volume)
  • Original equilibrium: $N<em>2O</em>4(g) \rightleftharpoons 2NO<em>2(g) \quad K</em>c = 0.21$
  • Increase pressure: $N<em>2O</em>4(g) \rightleftharpoons 2NO<em>2(g) \quad Q = \frac{[NO</em>2]^2}{[N<em>2O</em>4]} = 0.427$
  • Q > Keq, reaction has to go backwards to reach equilibrium.
• Change in Temperature
  • Is the reaction endothermic or exothermic as written? That matters!
    • Endothermic: Heat acts like a reactant; adding heat drives a reaction toward products.
    • Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants.
• The Haber process for producing ammonia from the elements is exothermic.
• Let’s give it a try
  • What can you do to shift the equilibria to (a) favor reactants or (b) favor products
  • Options: you can add or remove products and reactants; change the temperature; change the pressure)
    • $CaO(s) + CO<em>2(g) \rightleftharpoons CaCO</em>3(s) \quad DH_{rxn} = -179 kJ/mol$
      • Increase pressure = add more CO2, decreasing temperature
      • Decrease pressure = remove CO2, increasing temperature
    • $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$
      • Increase pressure = add more N2 or H2, removing NH3
      • Decrease pressure = remove N2 or H2, adding NH3

Additional info

• Solubility Equilibria
• The equilibrium constant expression is called the solubility-product constant. It is represented as $K_{sp}$.
• For example: $BaSO<em>4(s) \rightleftharpoons Ba^{2+}(aq) + SO</em>4^{2-}(aq)$
  • The equilibrium constant expression is $K<em>{sp} = [Ba^{2+}][SO</em>4^{2-}]$
• Factors Affecting Solubility
  The common-ion effect
  If one of the ions in a solution equilibrium is already dissolved in the solution, the solubility of the salt will decrease