Question

What percentage of reactant will remains after 120 s if the half-life is 60 s?

Answer 1

Given data: All - 605 t = 1205 Solution : Since order of reaction not given in question. we assume reaction is first order. til = 0.6932 K = 0.6932 – 0.6932 60 5 A 1/2 k= 0.0115551 According to kinetic equation of first order reaction k = 2.30 3 day where A = initial concentration of reactant Ao = remaining Concentration of reactant

above equation Substituting the value in 0.0 1155 st = 2303 log A 1205 I to 0.0115557x120 S 2-303 Ae lag A = 0.6018 taking Antilay of booth giale h = 2296416 h - 4 % AO= 40 x 100 = 14 x 100 = 25% Hence, 25% after of reactant will remains 1205.