17. The first-arderdeompositian of N20% a:325 K has a rate constuot of 1.20 10-3

Question

17. The first-arderdeompositian of N20% a:325 K has a rate constuot of 1.20 10-3-1, the initial creerta:ion of N2Osk 2 b, what is de concentration of N205 after 125 minutes? A) 0124 M D) 0355 M ) 0.174 M 8. The rate law for the reaction: H202+2++2o is rate= k(H202][-]·The following mechanism has been suggested. H2O2+1' HOI +0H- step t step 2 OH-+ H+ H2O Hot + H+ + I-12 +H20 step 3 dentify all insermediates incltuded in tris mechanism A. H+ and I- B. Ht and HOT C. HOI and OH- D. H only e. H0 and OH-

Explanation / Answer

Solution :-

Q 1.7 ) Rate constant 1.70*10^-3 s-1

Initial concentration = 2.88 M

Time = 12.5 min * 60 sec / 1 min = 750 sec

Final concentration = ?

Using the first order rate law equation lets calculate the final concentration

ln [A]t/[A]o = -k*t

ln [A]t/2.88 = -1.70*10^-3 s-1 *750 s

ln [A]t/2.88 = -1.275

[A]t/2.88 = anti ln [-1.275]

[A]t / 2.88 = 0.2794

[A]t = 0.2794 * 2.88 M

[A]t = 0.805 M

So the correct answer is option B

Q1.8) form the given reaction equations

The HOI and OH- are the intermediates

Therefore the correct answer is option C that is HOI and OH-

Q1.9 ) from the given graph energy verses reaction progress third step is the step which have higher energy of the product so the step 3 is the slow step

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