If the rate constant of a first order reaction is 4.606 10⁻³ ~s ⁻¹ , then find the time required for 400 ~g of the reactant to reduce to 50 ~g .

Step-by-step solution

Given, Rate constant, k=4.606 10⁻³ ~s ⁻¹ Concentration at, time t=50 ~g Initial concentration =400 ~g Now, formula, aligned k & = 2.303 t A₀ A_t t & = 2.303 4.606 10⁻³ 400 50 & = 2.303 0.00460 8 & = 2.303 0.9030 0.00460 & =452.08 ~s = 452.08 60 ~min t & =7.52 ~min aligned