2 Matching Annotations
  1. Last 7 days
    1. Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the decomposition reaction of N2O4N2O4\ce{N_2O_4} to NO2NO2\ce{NO2}. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:

      does this mean the relationship between the forward reaction and reverse are directly related to the k constants? or do we need to find the k constant first?

    1. The rate of dimerization of NO2NO2\ce{NO_2} (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N2O4N2O4\ce{N_2O_4} is zero, the rate of the dissociation reaction (forward reaction) at t=0t=0t = 0 is also zero. As the dimerization reaction proceeds, the N2O4N2O4\ce{N_2O_4} concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of N2O4N2O4\ce{N_2O_4} and NO2NO2\ce{NO_2} no longer change.

      I'm having a hard time understanding what this means. I thought being on equilibrium meant that both reaction had the same reaction rate regardless is the reaction was forward or backwards. so my question is will the reverse reaction will always be faster than the forward and if so how are these equation at equilibrium when they are happening at different rates. does equilibrium has to be at a particular time or position?