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Answer #1
![Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.](https://media.HomeworkLibcdn.com/media/df7/df77df35-671f-43ac-ac93-6e13b5a0055c/phptPzqpr.png)
![Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (](https://media.HomeworkLibcdn.com/media/179/179fc6ce-bb23-4b64-b986-86c44a9ac627/php1crIr0.png)
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
![Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.](http://img.homeworklib.com/questions/36036760-1432-11ec-bf83-5f338333e373.png?x-oss-process=image/resize,w_560)
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
Answer #1
Answen Reaction, 2 NO2 (g) + O₂(g) → N2O5 g) + O2 (g) Apply the rate law equation, Rate = K[ NO2]* COBJY I - 0 From Tridal 3.25x104 = K Co-osojme [0.800] Fromm Trial ② s.soxlot = K[o] [ 0.8oo]y Divide the equation ② by ② 5.50xlot = K[Hojte co.goo]% 3.25 x lot K[0-650] [0-800g e (1-69) = (1.69) de (1.69) (1.65) de First Order Kinetices with respect de - 1 to (NO2] From Trial 15.40x10r = K[1.76] [1.4001 -
Divided the equation ③ by 15.40 X = K[1-768 [ 1.40] 5.50 X 107 XC 1-10] [800]y 15.90 x104 1-76 (1.407 sosox lot 1.101 0.goo (2.8) = (1.6) (1.75) 4 (2.8)- = (2-8,4 : 14=1 First orden kinetic with respel to [03] put the value se and y eh Rate Law equaltia Rate = [NO2] ² [0371 | Rate = k [NO2] [03] Determinate of Rate constant (k), Froom Trial 6 K= Rete = 3.25X10 4 mst T NO 2] [03 ] (0.650) (0.800 m2 K= 3.25x104mst - -= 6.25xlot mitgt 0.52 M2 Ik = 6.25 Xlot misi
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Consider this initial-rate data at a certain temperature for the reaction described by NOCIo Initial rate of formation (M) of Cl2 (MIs) 0.600 1.51 x10 0.750 2.36 x10 0.900 3.40 x 10 Determine the value and units of the rate constant. Number Units M-1, s-1
Consider this initial-rate data at a certain temperature for the reaction described by [SO, Cl, 1, (M) Initial rate (M/s) SO,Cl2(g) → SO2(g) + Cl2(g) 1.80 x 10-6 Determine the order with respect to SOC12. 0.100 0.200 0.300 3.60 x 10-6 5.40 x 10-6 order: Determine the value and units of the rate constant. units: Initial rate data at a certain temperature is given in the table for the following reaction. 2 NOBr(g) —> 2 NO(g) + Br2(g) [NOBr), (M)...
Initial rate data at a certain temperature is given in the table for the following reaction. [NOCI), (M) 2 NOCI(g) — 2 NO(g) + Cl2(g) 0.600 Initial rate of formation of CI, (M/s) 1.40 x 10-5 2.19 x 10-5 3.16 x 10-5 0.750 0.900 Determine the value and units of the rate constant, k. Units:
Consider this initial-rate data at a certain temperature [NOBrlo Initial rate of formation for the reaction described by (M)of Br2 (M/s) 0.6001.08 x 10 0.750 0.9002.43 x 10 2 1.69 x 102 2 Determine the value and units of the rate constant. Number Units Select answer
estion 12 of 16 Consider this initial-rate data at a certain temperature for the reaction described by Initial rate C,H,Cl1O (M) C,H,Clg)C,H, (g) + HCl(g) (M/s) -30 0.805 x 10 0.100 Determine the order with respect to C, H, Cl. 0.200 1.61 x 10-30 0.300 2.42 x 10-30 order: Determine the value and units of the rate constant. k= Publisher: University Science Bo- Question Source: MRG-General Chemistry
Consider this initial-rate data at a certain temperature for the reaction described by 2NO_2 (g) + O_3 (g) rightarrow N_2 O_5 (g) + O_2 (g) Determine the value and units of the rate constant.
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