Chemical Kinetics
Top questions
See all →- The rate constant of a first-order reaction is \(6.93\times10^{-3}\) s-1. The time required for 75% completion is Answer: A. 200 s t_(1/2)=0.693/k=100 s; 75% completion takes two half-lives, i.e. 200 s. 1
- For the rate law \(r=k[\mathrm A]^2[\mathrm B]\), if [A] is doubled and [B] is halved, the rate becomes Answer: C. 2 times r'=k(2[mathrm A])²[mathrm B]/2=2r . 1
- The rate of a reaction doubles when temperature is raised from 300 K to 310 K. The activation energy is approximately (R = 8.314 J K-1 mol-1, log 2 = 0.301) Answer: C. 53.6 kJ mol-1 log2=E_a/2.303R·T₂-T₁/T₁T₂⇒ E_a=2.303×8.314×0.301×300×310/10≈53.6 kJ mol⁻¹. 1