Question

Using the data in the table, calculate the rate constant of this reaction. Trial [A] (M) 0.280 2 0.280 3 0.392 [B] (M) 0.320 0.864 0.320 Rate (M/s) 0.0169 0.123 0.0237 A+B - C +D Units

Answer 1

Q. using constant the of data this in the reaction. table , calculate Page 1 the rate A+B - C+D Trial (AJ (M) 0.280 To con T 0.320 Rate (MI) 0.0169 2 0.280 0.864 1 0.422 I 30.392 0.320 0.0237 Rate law equation (8) = K[A]" [8] we need to find a & y. In order to find a, we need to look at the data and determine what happens to "A' when I remains same (experiment 1 & 3). we notice that ' ľ become 1.4 times from ea epp 1 to exp 3. Exp-3 0-392 - 1.4 Exprd 0.280 and the rate also increasy by a factor of 1.4 Exp-3 0.0237 = 1.4 Expl 0.0169 Hence n= 1 look y for y, we need to at the data when, similarly A remains same & experiment tfal. we notice that 'B' times from epp I to becomes 2.7 exp 2 . Exp 2 0 .864 = 2.7 a factor of 7.27 Exp-1 + 0.320 and gate increased by EXD2 0.123 727 Expl → 0.0169

Page 2 hence, to to know to y (2.795 - 7:27 and y=2 Hence, we got our sate law equation, rak[A] [ 87² now, we have to find the sate constant (k), which I the conen of reactants to the rate of the reaction. we can plug data from any of the three experiments. so we choose expl. 0.0169 = K [0.280)' [ 0.320]? 0.0269 = KX 0:280 x 0.1024 k = 0.0169 0.0169 - berg 0.020 X 0.1024 010286 Rate constant = 0.59