Question

1. KClO₃ decomposes according to the reaction:           2KClO₃ → 2KCl + 3O₂
If the rate of decomposition of potassium chlorate at a certain time is determined to be 2.4 x 10^-2mol s^-1, what is the rate of formation of KCl at the same time?

2. The rate law for a reaction involving only A and B is 1^st order in A and 4^th order in B, respectively. If the concentration of A is increased by a factor of 4 and the concentration of B is tripled the new rate will be:

3. The reaction D ---> products is first order, and the rate constant is 2.7 x 10^-4 s^-1. If the initial concentration of D is 3.00 M, then what is the final concentration of D after 1.1 hr?

Answer 1

2KClO₃ → 2KCl + 3O₂

rate of reaction

-[KClO3]/2t = +[KCl]/2t

2.4*10^-2 /2        = +[KCl]/2t

1.2*10^-2         = +[KCl]/2t

+[KCl]/t = 2.4*10^-2 mole/sec

2. Rate law

   Rate = K[A][B]⁴

           = K*A/4*[3B]⁴

         = K[A][B]⁴ *81/4

81/4 times increases Or 20.25 ntimes rate inccreases

3. K = 2.303/t*log[A0]/[A]

t = 1.1hr = 1.1*60*60 = 3960 sec

[A0] = 3

K = 2.7*10^-4 sec^-1

2.7*10^-4 = 2.303/3960*log3/[A]

2.7*10^-4   = 5.815*10^-4 log3/[A]

0.4643     = log3/[A]

2.912    = 3/[A]

[A] = 1.03 M