Question

A certain first-order reaction has a rate constant of 0.007801/min at 300 K. What is the half-life (in minutes) of this reaction? Question 2 A certain first-order reaction with a single reactant has a rate constant equal to 0.0751/s at 1000 K. If the initial reactant concentration is 0.150M, how many seconds does it take to decrease to 0.0250M ? Question 3 1pts What data should be plotted to show that experimental concentration data fits a second-order reaction? 1/ [reactant] vs. time [reactant] vs. time In[reactant] vs. time

Answer 1

Question 1We know that k = 0.693/t₁/2t₁/2 = 0.693 / kHalf-life equation for a first-order reactionWhere k = 0.007801/mint₁/2 = 0.693/0.007801= 88.68 minutesAnswer: Half-life of this reaction = 88.68 minutes.Question 2We know that integrated rate law for first-order reaction is given as [A] = [A₀]e^(-kt) [A₀] = 0.150 M[A] = 0.0250 M = final concentrationk = 0.0751 / sWe need to find t where t is the time taken to decrease the concentration from 0.150 M to 0.0250 M. Let's plug in the given values to the equation.[A] = [A₀]e^(-kt)0.0250 M = 0.150 M e^(-0.0751t)Dividing by 0.150 M on both sides0.1667 = e^(-0.0751t)Taking natural logarithm of both sidesln 0.1667 = -0.0751 tln 0.1667/(-0.0751) = t.t = 11.1 s. (approximately)Answer: It takes 11.1 seconds to decrease the concentration from 0.150 M to 0.0250 M.Question 3Experimental concentration data fits a second-order reaction when plotted as 1/ [reactant] vs. time. Therefore, option A, 1/ [reactant] vs. time should be plotted to show that experimental concentration data fits a second-order reaction.