For a reversible reaction R P , at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants k_f and k_b, respectively. At time zero, the concentration of R is [ R ]_0 and the concentration of P is zero. At any given time, [ R ] and [ P ] are the concentrations of R and P , respectively. If k_b = 4k_f, the correct graphical representation of the reaction is
C.
For the reversible first-order reaction R P , at equilibrium, the rate of the forward reaction equals the rate of the backward reaction. k_f [ R ]_ eq = k_b [ P ]_ eq Given k_b = 4k_f, we can substitute this into the equilibrium condition: k_f [ R ]_ eq = 4k_f [ P ]_ eq [ R ]_ eq = 4[ P ]_ eq From the conservation of mass, the total concentration remains constant at any time t: [ R ]_ eq + [ P ]_ eq = [ R ]_0 Substituting [ R ]_ eq = 4[ P ]_ eq into the mass balance equation gives: 4[ P ]_ eq + [ P ]_ eq = [ R ]_0 5[ P ]_ eq = [ R ]_0 [ P ]_ eq = 0.2[ R ]_0 Using this to find [ R ]_ eq : [ R ]_ eq = 4 0.2[ R ]_0 = 0.8[ R ]_0 Therefore, as t , the concentration ratio [ P ] [ R ]_0 approaches 0.2 and [ R ] [ R ]_0 approaches 0.8. This is correctly represented in the third graph. Answer:
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