Question

For the reaction D → A + C when 1 is plotted versus the time in seconds, a straight line is obtained whose slope is 0.042 M-1 s-1. What is the concentration of D (in M) after 63.0 s if [Do = 0.50 M2 Hint given in feedback Answer: For the reaction A → B + C when A is plotted versus the time in minutes a straight line is obtained whose slope is-0013 M/min. What is the concentration of A (in M) after 17.00 min if Ao = 0.70 M? Answer: For the reaction A → B + C, when the natural log of A is plotted versus the time in seconds a straight line is obtained whose slope is -0.054 s1. What is the concentration of A (in M) after 45.0 s if [A]o = 0.10 M? Answer:

Answer 1

From the data, a straight line is obtained when a graph is drawn between Thus, the reaction is of second order Integrated rate law for a second order reaction is as follows: and time D] [D slope of the line is k Thus, k 0.042 M1s-1 Concentrati on of D at 63.0 sis calculated as follows D 0.50M 4.646 M D) D]-0.21M From the data, a straight line is obtained when a graph is drawn between [A] and time Thus, the reaction is of zero or der Integrated rate law for a second order reaction is as follows [A]=[A,-kt slope of the line is -k Thus, -k-0.013 M/min 0.013 M/min Concentrati on of A at 17.00 min is calculated as follows [A]-0.70 M-(0013 Mmin (17.00 min) [A]-0.48 M From the data, a straight line is obtained when a graph is drawn between log Aand time Thus, the reaction is of zero order Integrated rate law for a second order reaction is as follows: slope of the line is -k Thus,-k=-0.054 s-1 に0.054 s-1 Concentrati on of A at 45.0 sis calculated as follows log[A]=log(0.10 M)-(0.054 s-ı)(45.0 s) log [A]=-1.43 A] 10-143 [A]= 0.037 M