2006-09-13 10:27:43 · 6 answers · asked by nassafellow 2 in Science & Mathematics ➔ Chemistry
M^-3 s-1
2006-09-13 22:47:10 · answer #1 · answered by dan 2 · 2⤊ 0⤋
Mol^4 S^-4 But it would be a very slow reaction. 4th order means 4 molecules colliding simultaniously all with the correct energy and all corectly orientated for the reaction.
2006-09-14 08:53:17 · answer #2 · answered by christopher N 4 · 0⤊ 2⤋
sec^-1 M^-3
2006-09-13 10:38:23 · answer #3 · answered by TheOnlyBeldin 7 · 2⤊ 0⤋
Tripling the ClO2 concentration from 0.020 to 0.060 motives the cost to spice up through 0.0248/0.00276 = 9.0 situations, even as conserving [OH-] consistent. R is proportional to [ClO2]^n 9R is propotional to [3ClO2]^n, so n = 2 by reality 3^2 = 9. the cost is second order with take excitement in to [ClO2] Tripling OH- from 0.030 to 0.090 motives the cost to spice up through 0.00828/0.00276 = 3, even as conserving [ClO2] consistent. R is proportional to [OH-]^m 3R is proportional to [3OH-]^m, so m = a million by reality 3^a million = 3. the cost is first order with take excitement in to [OH-] . the cost regulation is R = ok[ClO2]^2[OH-]^a million employing trial a million information: ok = R / {[ClO2]^2[OH-]} = 0.0248 / {(0.060)^2(0.030)^a million} = 230 a million/M^2 R = 230 a million/s M^2 x (0.one hundred)^2 x (0.030)^a million = 0.069 M/s
2016-11-26 21:53:18 · answer #4 · answered by ? 4 · 0⤊ 0⤋
sec-1
2006-09-13 10:32:24 · answer #5 · answered by fatma m 2 · 0⤊ 3⤋
k = moldm^-3s^-1/ moldm^-3 x moldm^-3 x moldm^-3 x moldm^-3
= s^-1/ moldm^-3 x moldm^-3 x moldm^-3
= mol^-3dm^9s^-1
2006-09-15 11:52:44 · answer #6 · answered by Anonymous · 2⤊ 0⤋