Answer Happy

Learning Goal: For a general chemical equation A+B=C+D the equilibrium constant can be expressed as a ratio of the concentrations: [C] [D] AB Ke = If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows: forward rate = reverse rate = where ke and kr are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal: kf [A] = kr [C] [D] Thus, the rate constants are related to the equilibrium constant in the following manner: [C][D] [AB] K. = - ke [A] kr [C] [D] kr =
For a certain reaction, Ke = 6.09 and kf = 4.53x105 M2.s¹. Calculate the value of the reverse rate constant, kr. given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M 2.87 -1 include (multiplication dot) between each measurement. View Available Hint(s) kr = Submit ▾ Part B μÀ Value kr = . For a different reaction, Ke=6.72x109, kf = 6.56 x 105-1, and kr = 9.76x10-5s-1. Adding a catalyst increases the forward rate constant to 2.53x108 s 1. What is the new value of the reverse reaction constant, kr. after adding catalyst? HA Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M 2.87 include (multiplication dot) between each measurement. -1 ▸ View Available Hint(s) Value Units . SWD ? Units ?
Yet another reaction has an equilibrium constant Ke= 4.32 x 105 at 25 °C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 °C, what will happen to the equilibrium constant? View Available Hint(s) The equilibrium constant will O increase. decrease. not change.