Question

1. A certain reaction obeys second-order kinetics with respect to its only reactant (call it R) and has a rate constant of 0.0520 L mole^-1 s^-1.  If the initial concentration of reactant R is 0.265 moles/liter, what is the concentration of R after 3.50 minutes?                                                                                                             

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Answer 1

Answer:

The given reaction is

              2R Products

The rate law is -dR/dt = k [R]²

The integrated form of the above rate law is

[ 1/R_t - 1/R₀] = kt

Where R₀ is the initial concentration of R

          R_t is the concentration of R at time t

Given R₀ = 0.265 mol L^-1

         t = 3.5 min = 3.5x 60 = 210 s

         k = 0.0520 L mol^-1s^-1

Therefore 1/R_t = 1/R₀ + kt

                     = [1 / 0.265 mol L^-1 ]+ 0.0520 L mol^-1s^-1 x 210 s

                     = 3.774 L mol^-1 + 10.92 L mol^-1

                     = 14.694 L mol^-1

              R_t = [1 / 14.694] mol L^-1

                  = 0.068 mol L^-1

Thus the concentration of R after 3.5 minutes is 0.068 mol L^-1