Question

U KINETILS ANU ELUILIBRIUM Deducing a rate law from the change in concentration over time - Dariana A chemistry graduate student is studying the rate of this reaction: NH,OH(aq) NH3(aq) +H20(aq) She fills a reaction vessel with NH OH and measures its concentration as the reaction proceeds: time (seconds) NH,OH 0 0.600 M 10. 0.184 M 20. 0.109 M 30. 0.0771 M 40. 0.0597 M Use this data to answer the following questions. Write the rate law for this reaction. rate = k 0.0 Calculate the value of the rate constant k. x 6 ? Round your answer to 2 significant digits. Also be || sure your answer has the correct unit symbol.

Answer 1

Expression for the rate constant 14 order Kinetics: k = - In Expression for the rate constant 2010 order Kinetics: k = : **[4] [47] The given reaction belongs to that order which ever expression gives the constnat values of k. If the 14 expression gives the constant values of k, then the reaction is of 1st order. If the 2d expression gives the constant values of k, then the reaction is of 2nd order - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Initi al concentration of NH,OH, [A] =0.600 M Concentration of NH OH after 10 s, [A] = 0.184 M According to the 1st order kinetics, 1. [4]. - 1 By 0.600 M In A = 10 51" 0.184 M) = 0.1185-1 Concentration of NH, OH after 20 s, [A] = 0.109 M According to the 1st order kinetics, 1, [A] 1 0.600 M -ln {"LA [4] 20 s 0.109 M) Concentration of NH, OH after 30 s, [4] =0.0771 M According to the 1st order kinetics, k, = - 20.In = 0.0853 -1 LA 1 0.600 M =0.0684 51 Since the values of k does not comes out to be constant, this reaction is not of the 1st order. --------- -- -------- ------ ------------ ---- Initial concentration of NH,OH, [A], =0.600 M Concentration of NH,OH after 10 s, [A] = 0.184 M According to the 1st order kinetics, 1 1 1 1 1 1 1 = 0.377 M-1-1 *[[A] [A] 105 0.184M 0.600 M) Concentration of NH OH after 20 s, [A] = 0.109 M According to the 1st order kinetics, 111 17.1 1- 1 ) = 0.375 M-*s- t[A] [A] 20 $0.109M 0.600 M) = Concentration of NH, OH after 30 s, [A] = 0.0771 M According to the 1st order kinetics, [ 1 1] 1 1 = 0.377 M-1-1 t[[A] [A]. 30 $ 10.0771M 0.600 M) Concentration of NH, OH after 40 s, [A] = 0.0597 M According to the 1st order kinetics, 1 1 1 1 1 1 1 = 0.377 M-1-1 |[4] [A] 40 s (0.0597M 0.600 M) Since the values of k comes out to be constant, this reaction is of the 2nd order. Thus, rate law is as follows: Rate = k[NH,OH| Rate constant of the reaction,k= - ka+ka+kg+ka (0.377 M-\s-1)+(0.375 M+s4) +0.377 M--1) +0.377 M-1-1) = 0.376 M-:--