Question

Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical HO2, which decomposes into H,O2 and O2. The following data was obtained at 298 K [HO] (M) 8.50 5.10 Time (s) 0.00 0.60 1.00 1.40 1.80 2.40 3.60 2.60 1.80 1.10

Answer 1

HO, H₂O₂ + O₂ Given y CHon] (um) Time (Ms) 8.5 0.00 0,6 Be sub 2.6 109 1.8 1.1 2.4 We initially assume preaction भा от is 1st order as only reactant is involved, so reaction can't be Zero order.

Now, Assuming 1st order reaction, ly (1020) =.kt [H02] i 1 & when, to 0.6Ms. (Hoz] tous les 5.1.14 MM 0.6 85 KX 0.6 5.1 Msi R= 0.8513 i) when, to ims, [H02), = 36 AM em с кX/ K = 0.8591 Ms 2:6 MM (H02 Hi) when t= 109.4s, 119. lul 8:5 2.6 = 109 XK » K = 0.896 = 0.85 MB (1H02] 18. ? = 108. MM t=1.8ms, en 8) = 108xk 0.86 MS! → K

2- s en st) = 24** when t = 24 Ms - K= 0.85.19 MS 1 . Ex=rate constant rate Conget issumption fast-/ to, Constant kis almost constant at each Point of time This indicates that, our is correcta Reaction is 1st order reaction. " Rate Law: Rate = x (Hoz Post 2 Rate constant K= Average of k in (1).0;'), ) Liv Ivy experiments, 008513 + 0.8591+ 0.85 f. 01867 0.8519 5 - 0.8.59 MS Answer is 0.859 M5)