Law Of Mass Action Mcat

In reversible chemical reactions, a point called dynamic equilibrium is reached where the rate of products forming from reactants is equal to the rate of reactants forming from products. Unlike irreversible reactions, reversible reactions continue to take place, and the concentrations of products and reactants do not experience any net change at dynamic equilibrium. At this point, entropy is at its maximum, and change in Gibbs free energy (delta G) is zero.

Rate Processes in Chemical Reactions – Kinetics and Equilibrium

Dependence of reaction rate upon concentration of reactants; rate law

• The rate law is the equation that describes the rate = the product of reactants raised to some exponents.
• aA + bB → cC + dD
  • If the above reaction is single-step, then rate = k[A] a [B] b
  • If the above reaction is the rate-determining step of a multi-step reaction, then the rate of the multi-step reaction = k[A] a [B] b
  • If the above reaction is a multi-step reaction, then rate = k[A] x [B] y , where x and y are unknowns that correspond to the rate-determining step.

  • [A] (M)	[B] (M)	[C] (M)	rate (M/s)
    1	1	1	1
    2	1	1	4
    1	2	1	2
    1	1	2	1

  • r = k[A] x [B] y [C] z
  • From this table, a 2x increase in [A] corresponds to a 4x increase in the rate. 2 x = 4, so x = 2.
  • A 2x increase in [B] corresponds to a 2x increase in the rate. 2 y = 2, so y = 1.
  • A 2x increase in [C] corresponds to 1x (no change) in rate. 2 z = 1, so z = 0.
  • r = k[A] 2 [B] 1 [C] 0
  • r = k[A] 2 [B]
  • The k in the rate law is the rate constant.
  • The rate constant is an empirically determined value that changes with different reactions and reaction conditions.
  • Reaction order = sum of all exponents of the concentration variables in the rate law.
  • Reaction order in A = the exponent of [A]

  • Reaction Type	Reaction Order	Rate Law(s)
    Unimolecular	1	r = k[A]
    Bimolecular	2	r = k[A] 2 , r = k[A][B]
    Termolecular	3	r = k[A] 3 , r = k[A] 2 [B], r = k[A][B][C]
    Zero order reaction	0	r = k

  Rate determining step

  • The slowest step of a multi-step reaction is the rate determining step.
  • The rate of the whole reaction = the rate of the rate determining step.
  • The rate law corresponds to the components of the rate determining step.

  Dependence of reaction rate on temperature

  • Activation energy
    • Activated complex or transition state
      • Activated complex = what’s present at the transition state.
      • In the transition state, bonds that are going to form are just beginning to form, and bonds that are going to break are just beginning to break.
      • The transition state is the peak of the energy profile.
      • The transition state can go either way, back to the reactants, or forward to form the products.
      • You can’t isolate the transition state. Don’t confuse the transition state with a reaction intermediate, which is one that you can isolate.
      • The activation energy is the energy it takes to push the reactants up to the transition state.
      • ΔH is the difference between the reactant H and the product H (net change in H for the reaction).
      • H is heat of enthalpy.
      • Exothermic reaction = negative ΔH
      • Endothermic reaction = positive ΔH
      • k = Ae – Ea /_RT
      • k is rate constant, Ea is activation energy, T is temperature (in Kelvins), R is universal gas constant, A is a constant.
      • What this equation tells us: Low Ea, High T → large k → faster reaction.
      • When activation energy approaches zero, the reaction proceeds as fast as the molecules can move and collide.
      • When temperature approaches absolute zero, reaction rate approaches zero because molecular motion approaches zero.

      Kinetic control versus thermodynamic control of a reaction

      • A reaction can have 2 possible products: kinetic vs thermodynamic product.
        • Kinetic product = lower activation energy, formed preferentially at lower temperature.
        • Thermodynamic product = lower (more favorable/negative) ΔG, formed preferentially at higher temperature.
        • A reaction will occur if ΔG is negative.
        • ΔG = ΔH – TΔS
          

          Factors favoring a reaction	Factors disfavoring a reaction
          Being exothermic (-ΔH)	Being endothermic (+ΔH)
          Increase in entropy (positive ΔS)	Decrease in entropy (negative ΔS)
          Temperature is a double-edged sword. High temperatures amplify the effect of the ΔS term, whether that is favoring the reaction (+ΔS) or disfavoring the reaction (-ΔS)

        • A reaction will occur faster if it has a lower activation energy.

        Catalysts; the special case of enzyme catalysis

        • Catalysts speed up a reaction without getting itself used up.
        • Enzymes are biological catalysts.
        • Catalysts/enzymes act by lowering the activation energy, which speeds up both the forward and the reverse reaction.
        • Catalysts/enzymes alter kinetics, not thermodynamics.
        • Catalysts/enzymes help a system to achieve its equilibrium faster, but does not alter the position of the equilibrium.
        • Catalysts/enzymes increase k (rate constant, kinetics), but does not alter Keq (equilibrium).

        Equilibrium in reversible chemical reactions

        • Law of Mass Action
          • The Law of Mass Action is the basis for the equilibrium constant.
          • What the Law of Mass Action says is basically, the rate of a reaction depends only on the concentration of the pertinent substances participating in the reaction.
          • Using the law of mass action, you can derive the equilibrium constant by setting the forward reaction rate = reverse reaction rate, which is what happens at equilibrium.
            • For the single-step reaction: aA + bB cC + dD
            • r_forward = r_reverse
            • k_forward[A] a [B] b = k_reverse[C] c [D] d
            • k_forward /_kreverse = [C] c [D] d /_{[A] a [B] b}
            • Keq = [C] c [D] d /_{[A] a [B] b}
            • This holds true for single and multi-step reactions, the MCAT will not ask you to prove why this is so.
            • There are 2 ways of getting Keq
              • From an equation, Keq = [C] c [D] d /_{[A] a [B] b}
              • From thermodynamics, ΔG° = -RT ln (Keq)
                • Derivation: ΔG = 0 at equilibrium.
                • ΔG = ΔG° + RT ln Q
                • 0 = ΔG° + RT ln Q_{at equilibrium}
                • ΔG° = -RT ln Q_{at equilibrium}
                • ΔG = 0
                • r_forward = r_backward
                • Q = Keq
                • If Keq is much greater than 1 (For example if Keq = 10 3 ), then the position of equilibrium is to the right; more products are present at equilibrium.
                • If Keq = 1, then the position of equilibrium is in the center, the amount of products is roughly equal to the amount of reactants at equilibrium.
                • If Keq is much smaller than 1 (For example if Keq = 10 -3 ), then the position of equilibrium is to the left; more reactants are present at equilibrium.
                • If Q Keq, then the reaction is too far right, and is moving back left in order to reach equilibrium.
                • LeChatelier’s principle: if you knock a system off its equilibrium, it will readjust itself to reachieve equilibrium.
                • A reaction at equilibrium doesn’t move forward or backward, but the application of LeChatlier’s principle means that you can disrupt a reaction at equilibrium so that it will proceed forward or backward in order to restore the equilibrium.

                • Reaction at equilibrium	What will induce the reaction to move forward	What will induce the reaction to move backward
                  A (aq) + B (aq) C (aq) + D (aq)	Add A or B. Remove C or D.	Remove A or B. Add C or D.
                  A (s) + B (aq) C (l) + D (aq)	Add B. Remove D. Adding or removing solids or liquids to a reaction at equilibrium doesn’t do anything that will knock the system off its equilibrium. So, altering A and C won’t make a difference.	Remove B. Add D.
                  A (s) + B (aq) C (l) + D (g)	Add B. Remove D. Remove (decrease) pressure.	Remove B. Add D. Add (increase) pressure.
                  A (s) + B (g) C (l) + D (g)	Add B. Remove D. Since both side of the balanced equation contains the same mols of gas products, modifying pressure is of no use.	Remove B. Add D.
                  A (s) + B (aq) C (l) + D (aq) ΔH < 0	Add B. Remove D. Removing heat by cooling the reaction.	Remove B. Add D. Add heat by heating the reaction.

                Relationship of the equilibrium constant and standard free energy change

                • ΔG = ΔG° + RT ln Q
                  • Set ΔG = 0 at equilibrium.
                  • Q becomes Keq at equilibrium.

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                  Equilibrium in Reversible Chemical Reactions

                  In reversible chemical reactions, a point called dynamic equilibrium is reached where the rate of products forming from reactants is equal to the rate of reactants forming from products. Unlike irreversible reactions, reversible reactions continue to take place, and the concentrations of products and reactants do not experience any net change at dynamic equilibrium. At this point, entropy is at its maximum, and change in Gibbs free energy (delta G) is zero.

                  Two important concepts in reversible reactions are the equilibrium constant (Keq) and the reaction quotient (Qc). Keq describes the ratio of the concentration of products to the concentration of reactants at equilibrium, and the law of mass action states that this ratio is always constant if the temperature is constant. Qc uses the same equation as Keq but considers the concentrations of products and reactants at a certain snapshot in time (which, when Qc is being used, is typically not an equilibrium state). By comparing the current Qc to a reaction’s Keq at the same temperature, it is possible to determine if a reaction will proceed faster in the forward or reverse direction to reach equilibrium. If Keq is greater than Qc, the reaction will move in the forward direction, while if Qc is greater than Keq, the reaction will move in the reverse direction.

                  • Irreversible vs. reversible reactions
                    • In irreversible reactions, all reactants converted to products
                    • Dynamic equilibrium in reversible reactions
                      • Entropy at maximum and Gibbs free energy at zero when dynamic equilibrium is reached
                      • Keq equation for a hypothetical reversible reaction: [C]^c[D]^d/[A]^a[B]^b (where C and D are products with stoichiometric coefficients c and d, and A and B are reactants with stoichiometric coefficients a and b), using the concentrations of A, B, C, and D at equilibrium
                      • Exclusion of pure solids and pure liquids in equilibrium calculations
                      • Calculate Qc using the same equation as Keq
                      • Compare Qc to Keq at the same temperature to determine reaction progress
                      • Direction of the reaction based on Qc and Keq
                      • If Keq > Qc, the reaction will move in the forward direction; if Qc > Keq, the reaction will move in the reverse direction

                      

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                      What is the difference between dynamic equilibrium and static equilibrium in reversible chemical reactions?

                      Dynamic equilibrium refers to a state in reversible chemical reactions where the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of reactants and products. However, the reactants and products are continuously being converted into each other. In contrast, static equilibrium is a state where there is no net change in the system, and both the forward and reverse reactions have ceased.

                      How is Gibbs free energy related to equilibrium in reversible chemical reactions?

                      Gibbs free energy (ΔG) is an important thermodynamic concept that helps to predict the spontaneity of a chemical reaction and its relationship with equilibrium. At equilibrium, ΔG is equal to zero, indicating that the reaction is at its most stable state and no further net change will occur. If ΔG is negative, the reaction proceeds spontaneously in the forward direction, and if ΔG is positive, the reverse reaction is favored.

                      How does the law of mass action relate to equilibrium in reversible chemical reactions?

                      The law of mass action is a key principle governing the behavior of reversible chemical reactions at equilibrium. It states that the rate of a chemical reaction is proportional to the product of the molar concentrations of the reactants, each raised to the power of their stoichiometric coefficients in the balanced equation. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of reactants and products. The law of mass action leads to the concept of the equilibrium constant (Keq), which is calculated as the ratio of the concentrations of products to reactants, each raised to their stoichiometric coefficients.

                      What is the difference between the equilibrium constant (Keq) and the reaction quotient (Q)?

                      The equilibrium constant (Keq) is a dimensionless quantity that represents the ratio of the concentrations of products to reactants, each raised to their stoichiometric coefficients, when a reversible chemical reaction has reached equilibrium. It is a characteristic constant for a particular reaction at a specific temperature and does not change unless the temperature changes. The reaction quotient (Q), on the other hand, is a similar ratio of the concentrations or activities of products and reactants at any point during the reaction. Comparing Q to Keq helps to predict the direction in which the reaction will proceed. If Q < Keq, the forward reaction is favored and will proceed until equilibrium is reached. If Q >Keq, the reverse reaction is favored. When Q = Keq, the reaction is already at equilibrium.