Question

estion 4 La siguiente reacciones de primer orden, CHE - 2CH. Si la constante de rapidez es igual a 5.5 x 104 1 a 1000 K, cuinto tompo tardarin 0.35 mol de C2He en un contenedor de 1.00L para disminuir a 0.20 mol en el mismo contenedor? The following reaction is first order, CHE -- 2 CH3. If the rate constant is equal to 5.5 x 10- 4 1 at 1000K, how long wil it take for 0.35 mol of CzHe in a 1.00 L container to decrease to 0.20 mol in the same container? 17 min OOO 65 min 53 min

Answer 1

The first order Reaction is :

C2H6 -----> 2CH3

Given : Rate constant , k = 5.5 * 10^-4 s^-1

Initial concentration [C2H6]o = 0.35 mole/1.00L = 0.35 M

Concentration at time t, [C2H6] = 0.20mole/1.00L = 0.20 M

Now, using formula for 1st order reaction,

k = 2.303/t log(Ao/A)

where k is the rate constant, t is the time, Ao = initial concentration, A = concentration at time t

Thus, Ao = [C2H6]o = 0.35 molL^-1

and A = [C2H6] = 0.20 molL^-1

k = 5.5 * 10^-4 s^-1 , t = ?

Thus, putting the values, we get,

5.5 * 10^-4 s^-1 = 2.303/t log(0.35/0.20)

=> t = 2.303/5.5*10^-4 log(0.35/0.20) s

=> t = 1017.67 s

or t = 16.96 min

or t = 17 min

Hence, time taken will be 17min. Option-B is the correct answer.