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