Question

thee rate constant of production of 2B by the reaction A ---∆---> 2B is 2.48 × 10^-4 s^-1. A1:1 molar ratio of A and B in the reaction mixture is attained after ​

Answer 1

Answer:

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Explanation:

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).