The initial rate of a reation, A+B+C 2D+E, was measuredusing differentinitial co
ID: 683103 • Letter: T
Question
The initial rate of a reation, A+B+C 2D+E, was measuredusing differentinitial concentrations of A, B, and C. Results aresummarized in this table exprt. [A] [B] [C] Initial rate M/s 1 .100 .100 .100 7.3 x10-11 2. .200 .100 .100 1.2 x 10-9 3. .200 .200 .100 2.3 x 10-9 4. .200 .100 .300 1.1 x 10-8 a. What is the rate expression for the reation? b. What is the numerical value of k? C. What will be the initial rate in an experiment using thefollowing initial concentrations: [A]= .250 M; [B]=.300 M and [C]=.125 M? The initial rate of a reation, A+B+C 2D+E, was measuredusing differentinitial concentrations of A, B, and C. Results aresummarized in this table exprt. [A] [B] [C] Initial rate M/s 1 .100 .100 .100 7.3 x10-11 2. .200 .100 .100 1.2 x 10-9 3. .200 .200 .100 2.3 x 10-9 4. .200 .100 .300 1.1 x 10-8 a. What is the rate expression for the reation? b. What is the numerical value of k? C. What will be the initial rate in an experiment using thefollowing initial concentrations: [A]= .250 M; [B]=.300 M and [C]=.125 M? a. What is the rate expression for the reation? b. What is the numerical value of k? C. What will be the initial rate in an experiment using thefollowing initial concentrations: [A]= .250 M; [B]=.300 M and [C]=.125 M?Explanation / Answer
a. Rate expression: rate = k[A]x[B]y[C]z where, x, y ,z are orders A,B,C respectively. where, x, y ,z are orders A,B,C respectively. b. To find oreders with respect to A,B,C To find order withrespect to A: rate of expt2/rate of expt1 = { k[A]x[B]y[C]z}/{ k[A]x[B]y[C]z} 1.2 x 10-9 / 7.3 x 10-11 ={ k[0.2]x[0.1]y[0.1]z}/{ k[0.1]x[0.1]y[0.1]z}16.438 =(0.2/0.1)x 16.438 =(2)x take log on bothsides log16.438 = x log2 x = log16.438 / log2 = 4.01 x ˜4.0 similarly y =0.94 and z = 0.62 from expt 1 , 7.3 x 10-11 M/s = k(0.1)4.01(0.1)0.94(0.1)0.62 7.3 x 10-11 =k 2.69x10-6 k =2.7x10-5 c. initial rate =2.7x10-5*(0.250)4.01(0.300)0.94(0.125)0.62 = 9.24x10-9 M/s
take log on bothsides log16.438 = x log2 x = log16.438 / log2 = 4.01 x ˜4.0 similarly y =0.94 and z = 0.62 from expt 1 , 7.3 x 10-11 M/s = k(0.1)4.01(0.1)0.94(0.1)0.62 7.3 x 10-11 =k 2.69x10-6 k =2.7x10-5 c. initial rate =2.7x10-5*(0.250)4.01(0.300)0.94(0.125)0.62 = 9.24x10-9 M/s 7.3 x 10-11 =k 2.69x10-6 k =2.7x10-5 c. initial rate =2.7x10-5*(0.250)4.01(0.300)0.94(0.125)0.62 = 9.24x10-9 M/s
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