Question

1. The rate constant of the production of 2B(g) by the reaction A(g) 2B(g) is 2.48 x 10-4S-1











A 1: 1 molar ration of A to B in the reaction mixture is attained after :











a. 26.25 minutes b. 27.25 minutes c. 28.25 minutes d. 0 minute

Answer 1

Lets assume after time 't' the molar ratio becomes 1 : 1

So the reaction is,

                  A     →  2B

[Initially]      A₀        0

[After time t](A₀-x)   2x

Thus,

A₀-x = 2x

or, A₀ = 3x

or, x = A₀/3

Now from the equation it can be deducted that it's a first order equation and for that the rate equation is,

t = (2.303/k) log [A₀ / (A₀ - X)]

Given k = 2.48 X 10⁻⁴ s⁻¹

Thus, t = (2.303 / 2.48 X 10⁻⁴) log [A₀ / {A₀ - (A₀/3)}]

or, t = 9286.29 log [3/2]

or, t = 9286.29 X 0.176

or, t = 1635.234

Thus, required time is 1635.234 seconds or 27.25 minutes(b).

Answer 2

Answer:

t = 27.25 minutes

Explanation:

just see the uploaded photo