The half life of first order reaction is 850 s. The initial concentration of the reactant is 0.06 mol dm ⁻³ . What concentration would remain after 1200 s ?

Step-by-step solution

For a first order reaction, the rate constant k is given by: k = 0.693 t_ 1/2 Substituting the given half-life t_ 1/2 = 850 s: k = 0.693 850 = 8.15 10⁻⁴ s ⁻¹ The integrated rate law for a first order reaction is: k = 2.303 t ( [A]₀ [A]_t ) Substituting the values t = 1200 s and [A]₀ = 0.06 mol dm ⁻³ : 8.15 10⁻⁴ = 2.303 1200 ( 0.06 [A]_t ) ( 0.06 [A]_t ) = 8.15 10⁻⁴ 1200 2.303 0.425 Taking the antilog on both sides: 0.06 [A]_t = 10^ 0.425 2.66 [A]_t = 0.06 2.66 0.0226 mol dm ⁻³ Rounding off, the concentration remain