For the reaction, 2 A(g) + 2 B(g) => C(g) + 3 D(g), thefollowing data were obtai
ID: 691443 • Letter: F
Question
For the reaction, 2 A(g) + 2 B(g) => C(g) + 3 D(g), thefollowing data were obtained at constant temperature. Experiment Initial [A], mol/L Initial [B], mol/L Initial Rate, M/min 1 0.11 0.15 2.0 x 10-5 2 0.11 0.30 2.0 x 10-5 3 0.22 0.45 4.0 x 10-5 4 0.22 0.60 4.0 x 10-5 Calculate the value of k to 3 significant figures.CORRECT ANSWER: .000182
HELP!
CORRECT ANSWER: .000182
HELP! Experiment Initial [A], mol/L Initial [B], mol/L Initial Rate, M/min 1 0.11 0.15 2.0 x 10-5 2 0.11 0.30 2.0 x 10-5 3 0.22 0.45 4.0 x 10-5 4 0.22 0.60 4.0 x 10-5
Explanation / Answer
The given reaction is 2A(g) + 2 B(g) -----> C(g) + 3D(g) Experiment Initial [A] , mol/L Initial [B] , mol/L Initial Rate, M /min 1 0.11 0.15 2.0*10-5 2 0.11 0.30 2.0*10-5 3 0.22 0.45 4.0*10-5 4 0.22 0.60 4.0*10-5 so from the data given we know that , by keepingA is fixed and we will increase the concentration of B isdouble, the rate of change in B is not found . hence the order ofthe reactant of B is " Zero ". Similarly now we willincrease the concentration of A is double by following B alsodouble , then the rate of change is double ,hencethe order of A is " first ". sorate R = k[A]1[B]0 Now rate constant k = rate ,R /[A]1[B]0 Tocalculate rateconstant k we will consider an experiment 1 . here Initial [A] = 0.11 , Initial [B] = 0.15 andInitial Rate = 2.0*10-5 M /min Now rate constant k = (2.0*10-5 M /min ) /(0.11M)1*(0.15M)0 = 0.000182 min-1 Note : for firstorder reactions rate constant unit is min-1. = 0.000182 min-1 Note : for firstorder reactions rate constant unit is min-1.Related Questions
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