Question

13.What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 30.0 minutes? Assume first-order kinetics.

The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH3(g) → P(g) + 6H2(g) The half-life of the reaction is 35.0 s at 680°C. Calculate the first-order rate constant for the reaction: s−1 Calculate the time required for 88.0 percent of the phosphine to decompose: s

Answer 1

13. K = 2.303/t log(a/a-x)

    K   = 2.303/30 log(100/100-75)

          = 0.0767*0.6020 = 0.0462sec^-1

K   = 0.693/t1/2

t1/2 = 0.693/0.0462   = 15 sec

14. K = 0.693/t1/2

   K    = 0.693/35   = 0.0198sec^-1

. K = 2.303/t log(a/a-x)

    t = 2.303/0.0198 log(100/100-88)

         = 116.3*0.9208   = 107.09sec