CHEM 111 Unit 14

Introduction to Chemical Equilibrium

  • Definition:

    • Chemical equilibrium is the state in which the concentration of reactants and products remains constant over time.

    • The specific conditions for equilibrium are when the rate of the forward reaction equals the rate of the reverse reaction:

    extReactants<br>ightleftharpoonsextProductsext{Reactants} <br>ightleftharpoons ext{Products}

  • Misconceptions:

    • Equilibrium does not imply that the concentrations of reactants and products are equal.

    • Equilibrium also does not mean that no reactions are occurring; it indicates a dynamic state where reactions continue to occur but in balanced rates.

  • Dynamic Nature of Equilibrium:

    • The forward and reverse reactions occur simultaneously, leading to a dynamic equilibrium where all species (reactants and products) remain present.

    • Example of a dynamic equilibrium:

    N<em>2(g)+3H</em>2(g)<br>ightleftharpoons2NH3(g)N<em>2(g) + 3H</em>2(g) <br>ightleftharpoons 2NH_3(g)

    • All three species (N2, H2, NH3) remain present in dynamic equilibrium.

Chemical Equilibrium: Reaction Rates

  • Essential Concepts:

    • The equilibrium state is defined quantitatively by the rates of the reactions:

    • Forward reaction rate = Reverse reaction rate

    • This balance leads to a stable ratio of concentration between products and reactants.

Concentration in Chemical Equilibrium

  • Graphical Representation:

    • Concentation verses time graphs show reactants and products approaching constant values as equilibrium is reached.

    • Graph indicates that while the concentration of reactants decreases, the concentration of products increases until they stabilize at equilibrium.

The Haber Process: Example of Chemical Equilibrium

  • Haber Process:

    • Reaction:
      N<em>2(g)+3H</em>2(g)<br>ightleftharpoons2NH3(g)N<em>2(g) + 3H</em>2(g) <br>ightleftharpoons 2NH_3(g)

    • Steps to Consider:

    1. Initially, only nitrogen ($N2$) and hydrogen ($H2$) are present.

    2. As ammonia ($NH3$) is formed, the concentrations of $N2$ and $H_2$ decrease, reducing further reaction likelihood.

    3. Equilibrium concentrations become constant when formation and dissociation rates equalize.

Stoichiometry in Equilibrium Calculations

  • Example Problem:

    • Given the equilibrium amount of NH3 is 0.080 mol/L, calculate the equilibrium amounts of other species in the reaction:

    • Using stoichiometric ratios:

      • $2x$ = 0.080 mol/L for $NH_3$ yields:

      • $Eq. amt of N_2 = 1.000 - 0.040 = 0.960 mol/L$

      • $Eq. amt of H_2 = 3.000 - (3 imes 0.040) = 2.880 mol/L$

Equilibrium Constant (K)

  • Definition:

    • The equilibrium constant ( extit{K}) is defined for the equilibrium of a reaction:

    K=rac[extProducts][extReactants]K = rac{[ ext{Products}]}{[ ext{Reactants}]}

    • The concentrations are raised to the power of their respective stoichiometric coefficients from the balanced equation.

  • Mathematical Relationship:

    • For the reaction:
      aA+bB<br>ightleftharpoonscC+dDaA + bB <br>ightleftharpoons cC + dD

    • The equilibrium constant expression would be:

    K=rac[C]c[D]d[A]a[B]bK = rac{[C]^c[D]^d}{[A]^a[B]^b}

Example to Calculate Equilibrium Constant (Kc)

  • Given Data:

    • [CO] = 0.0613 M

    • [H2] = 0.1893 M

    • [CH4] = 0.0387 M

    • [H2O] = 0.0387 M

  • Calculate Kc:

    • The Kc expression for the reaction is formulated as:

    Kc=rac[CH4][H2O][CO][H2]Kc = rac{[CH4][H2O]}{[CO][H2]}

    • Plugging in concentrations:

    Kcext(aftercalculations)=3.93Kc ext{(after calculations)} = 3.93

Implications of K Values

  • Interpretations of K:

    • If $K >> 1$, equilibrium favors products.

    • If $K << 1$, equilibrium favors reactants where little product is formed.

    • If $K = 0$, no reaction occurs (only reactants remain).

    • If $K = ext{∞}$, reaction is irreversible with all products formed.

Equilibrium in Gas Phase Reactions

  • Using Partial Pressure:

    • Equilibrium constants ($K_p$) are derived from partial pressures instead of molar concentrations.

    • The relationship between $Kp$ and $Kc$ can be expressed as:

    K<em>p=K</em>c(RT)rianglengasK<em>p = K</em>c(RT)^{ riangle n_{gas}}

    • Where $ riangle n{gas} = n{products} - n_{reactants}$.

  • Example Problem:

    • Given $Kc$ = 2.8 x 10^2, calculate $Kp$ at 1000K considering:

    • $Kp = Kc imes (RT)^{ riangle n}$

Effects of Conditions on Equilibrium

  • Le Chatelier's Principle:

    • States: When a system at equilibrium is disturbed, it will shift in a direction that counteracts the change (e.g., changes in concentration, pressure, temperature).

  • Changes in Concentration:

    • If a reactant is added, the reaction proceeds towards products; if a product is added, the reaction will shift towards reactants.

  • Changes in Pressure/Volume:

    • When the total number of gaseous moles changes, the reaction shifts to reduce that change; if moles of products equal moles of reactants, pressure changes do not affect the equilibrium.

  • Effects of Temperature:

    • For exothermic reactions, higher temperatures decrease Kc.

    • For endothermic reactions, higher temperatures increase Kc.

  • Catalyst Effects:

    • Catalysts increase the rate of reaching equilibrium without affecting the equilibrium concentrations of products and reactants.