Question

decomposition of A is a first order reaction and the [A] is monitored. It was reported that [A] reduced from 0.100M to 0.00670M in 8 minutes. Determine i. The rate constant k ii. The half life iii. [A] when t= 5mins iv. t when [A] = 0.0100M ​

Answer 1

Answer:

Explanation:

For a first order reaction, the relation between concentration profile and time is given by:

ln (Cₐ₀/Cₐ) = kt

where,

Cₐ₀ = Initial concentration of reactant 'A'

Cₐ = final concentration of reactant 'A' at any time t

k = rate constant

t = time

From the question: Cₐ₀ = 0.100 M

                                Cₐ = 0.0067 M

                                 t = 8 minute

i) Putting the values in the above expression, we can calculate 'k' as follows

      ln (0.100/0.0067) = k x 8 min

    ⇒ 2.70 = k x 8 min

     ⇒  k = 0.3375 min⁻

ii) Half life of a first order reaction is given by:

t₀.₅ = 0.693/k

⇒t₀.₅ = 0.693/0.3375

⇒t₀.₅ = 2.053 min

iii) t = 5 mins

   Cₐ₀ = 0.100 M

    k = 0.3375 min⁻

Putting the values in the above expression

ln (0.100/Cₐ) = 0.3375 min⁻ x 5 min

⇒ln (0.100/Cₐ) = 1.6875

⇒0.100/Ca = 5.40

⇒Cₐ = 0.100/5.40

⇒Cₐ = 0.0185 M

iv) Cₐ = 0.0100 M

    k = 0.3375 min⁻

   Cₐ₀ = 0.100 M

Substituting the values:

ln (0.100/0.0100) = 0.3375 x t

⇒2.30 = 0.3375 x t

    t = 6.82 min