Question

[time] [Z] 4. Suppose the data shown are for the second order reaction: → products. What is the value of the rate constant k? O min 50.000 atm 6.250 3.333 2.273. 5. For a first order reaction (G - products) with k = 0.173 min', suppose a chemist runs the reaction starting with an initial concentration [G]. = 12.0 M. a. How many minutes will it take for [G] to decrease to 4.70 M? b. What [G] will remain after 2.93 min? 6. For a second order reaction (R - products) with k = 0.55 L. mol-.hr , suppose a chemist runs the reaction starting with an initial concentration [R), = 0.500 M. How many hours will it take for [R] to decrease to 0.200 M? b. What [R] will remain after 3.4 hr?

Answer 1

TuHe WU., Integrated Rate law expression for and order - I - iz Kt co (3)+ Zo _ __ Rate constant = t= time Ina - 20 = t = 6.250. 50.000 atm t= 2 min À - 2x.ke Ilala KX2 66250 6.250 50 50- 6.250 0.147? 312.5 0:07 atm's!! } ] Kz_ © Ka - Ke 0.173.mini' С т . ч. 46 м integrated rate law -kt = er [ho] (a), : 12.0M з for first order a t an 0.173 5:094 ln ( 12. 1 1 In [2.55] - 0.173 5 .42 min Iho t= 2.93 min (ht] = ? = 0.173 X 2.93 In [ho] Kt (6+) [907 - 0.51

Date - Page No.: e oosid e ( In a=:Y hoc от a ey — no 1.67 GT- = .7.19 M. ho 1.67 12 1.67 St. At - Rol [R]: 0.5M 2 [B+1= 0.2 k= 0.55 L malt hr- 1 0 :55+ 0.2 0.5 not = 1 {5-23 = 3 0.55 0.55 4 = 5 HS : _ || ( be t = 3.4 kv. [R] =? 0.558 3.4 [R] 0.5 1.87 + 2 (R2 [RO L _ 0:26 m. 13_ 3-87