What Does It Mean if Keq 1

Reaction rates

• The reaction rate is defined as the rate of change in the concentration of reactants or products. ie. how fast a reactant gets used upwardly, and how fast a production gets produced.
• Charge per unit = –^ΔReactant/_ΔTime
  = how fast a reactant disappears.
• Rate =
  ^ΔProduct/_ΔTime
  = how fast a production forms.
• The unit for rate is molarity per second, or M/due south.

Dependence of reaction rate upon concentration of reactants; charge per unit 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-pace, and then rate = thou[A]^a[B]^b
  • If the above reaction is the rate-determining pace of a multi-pace reaction, so the rate of the multi-step reaction = grand[A]^a[B]^b
  • If the higher up reaction is a multi-step reaction, then charge per unit = 1000[A]^x[B]^y, where x and y are unknowns that stand for to the rate-determining step.
• To determine the rate law, you refer to a table of rates vs reactant concentrations.

  • [A] (M)	[B] (M)	[C] (Yard)	rate (M/south)
    1	1	1	ane
    2	1	1	4
    1	2	i	2
    ane	1	2	ane

  • r = k[A]^x[B]^y[C]^z
  • From this table, a 2x increase in [A] corresponds to a 4x increase in the rate. 2¹⁰
    = 4, and then ten = 2.
  • A 2x increase in [B] corresponds to a 2x increase in the rate. two^y
    = 2, so y = 1.
  • A 2x increase in [C] corresponds to 1x (no modify) in rate. 2^z
    = ane, and then z = 0.
  • r = k[A]ⁱⁱ[B]ⁱ[C]
  • r = yard[A]²[B]
• rate constant
  • The k in the rate law is the rate abiding.
  • The rate constant is an empirically determined value that changes with different reactions and reaction conditions.
• reaction order
  • Reaction lodge = sum of all exponents of the concentration variables in the rate law.
  • Reaction social club in A = the exponent of [A]

  • Reaction Type	Reaction Order	Charge per unit Law(s)
    Unimolecular	i	r = grand[A]
    Bimolecular	2	r = thousand[A]two, r = grand[A][B]
    Termolecular	3	r = grand[A]iii, r = g[A]ii[B], r = thousand[A][B][C]
    Zero order reaction		r = k

Rate determining step

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

Dependence of reaction rate on temperature

• Activation energy
  • Activated complex or transition state
    • Activated circuitous = what’due south present at the transition land.
    • In the transition state, bonds that are going to form are simply beginning to grade, and bonds that are going to break are just beginning to break.
    • The transition state is the peak of the energy profile.
    • The transition country can become either fashion, back to the reactants, or forwards to form the products.
    • Yous can’t isolate the transition land. Don’t misfile the transition state with a reaction intermediate, which is one that y’all can isolate.
  • Interpretation of energy profiles showing energies of reactants and products, activation energy, ΔH for the reaction
    • 
    • The activation energy is the free energy it takes to push the reactants up to the transition state.
    • ΔH is the departure between the reactant H and the product H (net change in H for the reaction).
    • H is oestrus of enthalpy.
    • Exothermic reaction = negative ΔH
    • Endothermic reaction = positive ΔH
• Arrhenius equation

Kinetic command versus thermodynamic command of a reaction

• A reaction can have two possible products: kinetic vs thermodynamic product.
  • Kinetic production = lower activation free energy, formed preferentially at lower temperature.
  • Thermodynamic product = lower (more than favorable/negative) ΔG, formed preferentially at college temperature.
• 
• Thermodynamics tells you whether a reaction will occur. In other words, whether it is spontaneous or non.
  • A reaction will occur if ΔG is negative.
  • ΔG = ΔH – TΔS
    

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

• Kinetics tells you how fast a reaction will occur.
  • A reaction volition occur faster if it has a lower activation energy.

Catalysts; the special case of enzyme catalysis

• Catalysts speed upwardly a reaction without getting itself used up.
• Enzymes are biological catalysts.
• Catalysts/enzymes act past lowering the activation energy, which speeds up both the forward and the reverse reaction.
• Catalysts/enzymes alter kinetics, not thermodynamics.
• Catalysts/enzymes help a arrangement to attain its equilibrium faster, just does not alter the position of the equilibrium.
• Catalysts/enzymes increment m (rate constant, kinetics), but does not alter Keq (equilibrium).

Equilibrium in reversible chemical reactions

• Police force of Mass Activity
  • The Law of Mass Action is the ground for the equilibrium constant.
  • What the Law of Mass Action says is basically, the charge per unit of a reaction depends only on the concentration of the pertinent substances participating in the reaction.
  • Using the law of mass activeness, you can derive the equilibrium abiding by setting the forrard reaction rate = reverse reaction charge per unit, which is what happens at equilibrium.
    • For the unmarried-step reaction: aA + bB <–> cC + dD
    • r_forrad
      = r_reverse
    • k_frontwards[A]^a[B]^b
      = g_reverse[C]^c[D]^d
    • ^{thousand_forward}
      /_kreverse
      
      =
      ^[C]c[D]d
      /_[A]^a[B]^b
      
    • Keq =
      ^[C]c[D]d
      /_[A]^a[B]^b
      
    • This holds true for unmarried and multi-footstep reactions, the MCAT volition not ask y’all to bear witness why this is then.
• the equilibrium constant
  • 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}
    • At equilibrium:
      • ΔG = 0
      • r_forrard
        = r_backward
      • Q = Keq
  • Keq is a ratio of one thousand_forward
    over grand_backward
    • If Keq is much greater than 1 (For instance if Keq = 10³), then the position of equilibrium is to the right; more products are present at equilibrium.
    • If Keq = ane, then the position of equilibrium is in the center, the corporeality of products is roughly equal to the corporeality of reactants at equilibrium.
    • If Keq is much smaller than 1 (For example if Keq = x^-3), then the position of equilibrium is to the left; more reactants are present at equilibrium.
  • The reaction quotient, Q, is the same as Keq except Q can be used for whatsoever point in the reaction, not just at the equilibrium.
    • If Q < Keq, so the reaction is at a point where it is still moving to the right in order to reach equilibrium.
    • If Q = Keq, the reaction is at equilibrium.
    • If Q > Keq, then the reaction is too far right, and is moving back left in order to reach equilibrium.
  • The reaction naturally seeks to reach its equilibrium
• awarding of LeChatelier’south principle
  • LeChatelier’southward principle: if you lot 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’south principle means that you can disrupt a reaction at equilibrium and then that it will proceed forward or backward in order to restore the equilibrium.

  • Reaction at equilibrium	What volition induce the reaction to motility forrad	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 annihilation that will knock the system off its equilibrium. So, altering A and C won’t make a deviation.	Remove B. Add together D.
    A (due south) + B (aq) <–> C (l) + D (one thousand)	Add together B. Remove D. Remove (decrease) pressure level.	Remove B. Add D. Add (increase) pressure.
    A (s) + B (g) <–> C (l) + D (one thousand)	Add B. Remove D. Since both side of the balanced equation contains the aforementioned mols of gas products, modifying pressure level is of no use.	Remove B. Add D.
    A (southward) + B (aq) <–> C (50) + D (aq) ΔH < 0	Add B. Remove D. Removing heat by cooling the reaction.	Remove B. Add D. Add together estrus by heating the reaction.

Relationship of the equilibrium constant and standard free energy modify

• ΔG = ΔG° + RT ln Q
  • Fix ΔG = 0 at equilibrium.
  • Q becomes Keq at equilibrium.
• 0 = ΔG° + RT ln (Keq)
• ΔG° = -RT ln (Keq)