Ans 22 and 23 questions


22. The rate constant of a reaction of zero order in A is 3.0 × 10^–3 mol L^–1s^–1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M?

23.  For a reaction, X(g) → Y(g) + Z(g)
the half-life period is 10 min. In what period of time would the concentration of X be reduced to 10% of original concentration?

Dear Student,

Solution for Q22:
For zero order reaction:
Rate constant (k) = 0.0030 mol L ^–1 s^–1
Initial concentration [R]₀ = 0.10 M
Final concentration [R] = 0.075 M
Time = ?
For Zero order reaction rate constant is given by
So,
t = ( 0.10 - 0.075) /  0.0030
t = 0.025 / 0.0030 = 8.33 s
It will take 8.33 seconds for the initial concentration of A to fall from 0.10 M to 0.075 M.

Solution for Q23:
For the first-order reaction,
k = 0.693/t_1/2 
⇒ k = 0.693/10 = 0.0693 min^-1 

Also, t = 2.303/k*log(A_o/A)
And A_o 10% of A_o = 10/100*A_o = 0.1A_o
So,
t = 2.303/k*log(A_o/A)
⇒ t = 2.303/0.0693*log(1/0.1) = 33 min
Hence, the time period would be of 33 min.

Regards