Dynamic Equilibrium

Key Questions

  • How does concentration effect dynamic equilibrium?

    Answer:

    The Law of Mass Action states that when a system (e.g., chemical reaction) is at equilibrium, the ratio of products and reactant concentrations is equal to the equilibrium constant for that reaction. If we change the concentration of any species (e.g., by adding reactant to the mixture), then the reaction will shift toward reactants or products in order to restore the ratio of concentrations to the equilibrium value.

    Explanation:

    Example:
    When #NO_2# is in equilibrium with its dimer, #N_2O_4#, the ratio of concentrations (expressed in mol/L) is equal to 4.7 at room temperature.

    #2 NO_2 harr N_2O_4#

    #K_c=([N_2O_4])/([NO_2]^2)=4.7#

    So, if the concentration of #NO_2# is 0.02M, then we know that the equilibrium concentration of #N_2O_4# must be #4.7(0.02)^2=1.88times10^-3M#

    If the system is in a 1L container at constant volume and temperature and we add 0.03 mol of #NO_2#, then for a short time the concentration of #NO_2# is increased to 0.05M. The non-equilibrium ratio of concentrations is
    #Q=([N_2O_4])/([NO_2]^2)=(1.88times10^-3)/((0.05)^2)=0.75#
    which is smaller than the equilibrium ratio of 4.7. In order to restore equilibrium, the reaction shifts toward products, consuming #NO_2# and producing #N_2O_4# until the equilibrium rations are restored:

    #K_c=([N_2O_4])/([NO_2]^2)=(0.00188+x)/((0.05-2x)^2)=4.7#

    Solving this quadratic equation for the change in concentration, #x#, gives #x=5.37times10^-3M#. The new equilibrium concentrations will be
    #[N_2O_4]=0.00188+0.00537=0.00725M#
    #[NO_2]=0.05-2(0.00537)=0.0392M#
    and the new ratio is
    #([N_2O_4])/([NO_2]^2)=4.72#, which is equal to the equilibrium constant within the rounding error of the significant digits of the calculation.

    Chuck W. · 2 · Jun 18 2015
  • Why is chemical equilibrium dynamic?

    Answer:

    Good question....we can approach the answer on the basis of #"dynamic"# rates of reaction....

    Explanation:

    We write the general equilibrium as:

    #A+BrightleftharpoonsC+D#

    As with any reaction there is a rate forward, #"rate forward"=k_f[A][B]#, and of course a rate backward, #"rate backward"=k_r[C][D]#, i.e. dynamic processes...

    Now, by definition, the chemical condition of equilibrium is defined when the forward and reverse rates are equal:

    #"i.e." " rate forward "=" rate backward"#,

    #"i.e."# #k_f[A][B]=k_r[C][D]#

    And upon rearrangement,

    #k_f/k_r=([C][D])/([A][B])#

    And the quotient #k_f/k_r=K_c#, otherwise known as the equilibrium constant for the reaction. Generally #K_c# is a constant for a given temperature. #K_c# can be large (i.e. the products are favoured at equilibrium) or small (the reactants are favoured). #K_c# may be formally related to the thermodynamic properties of the reaction, though I am not going to do it here.

    Now #K_c# must be measured, and as a constant it cannot be altered. However, a chemist or engineer can certainly manipulate the equilibrium. For instance, if we remove (somehow) the products of the reaction, #C# and #D#, the equilibrium will have to re-establish itself, and it does this by moving to the right as written to satisfy the equilibrium equation, and to re-establish equilibrium concentrations of #C# and #D#. On the other hand, if we pump more reactant into the equilibrium, the equilibrium will move in a forward direction to cope with increased #[A]# and #[B]#.

    There should be many answers here that deal with equilibria. If there is a specific problem or query, ask, and someone will help you.

    anor277 · 2 · Mar 8 2018

Questions

Chemical Equilibrium