Question

Using the data in the table, calculate the rate constant of this reaction. | Trial (A) (M) 0.380 0.380 0.532 B (M) 0.240 0.648 0.240 Rate (M/s) 0.0117 0.0853 A+B C+D 0.0164 Units

Answer 1

SOLUTION The given reaction is A + B - C + By rate law we know that, Rate of reaction - K ATA [B]6) where k is the rate constant. [A] and [B] are the concentrations of A and B and a and to represents the order of reaction with respect to A and B - using trial 1 and toide 2 data, 0.0117= k (0.38039 (0.240)b and, 0.0853= k 10.3803a 10.6489b Dividing these two equatit On equating the value of ke we get 1.0-38019 / 0.64 0.08 53 0.0117 (0.380) 0.stol 7.29 - (vor (267) & 7.29 - (2.7 Jb [also, 7.29= 20727 u b=2 trial using t data, and trial 3 Doll4 - 10.532 19 10.24016 - 0.3800 10. 2400 lotho = (1.4)9 11b 8. a=1 ( I OD 117

Thus sate daw. for given reaction is, rate of reaction = K[A] [B] 2) values of trial 1, 0.0117 K 10.380] 10.2h0j? k = 0.53 45 M2 grl Using Rate constant - 0.5345 M2gt.

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