Need help knowing how we got these answers 8. A series of experiments were perfo

Question

Need help knowing how we got these answers 8. A series of experiments were performed for the reaction below: Ch072-(ag) + 6 1-(w) + 14H+(ay) 2 Cr"(ay) + 3 12(ag) + 7 H20() for which the following initial instantaneous rates of reaction were obtained initial CrOnitial MinitialH initial rate (M/s) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 initial (Cr O 0.0040 0.0040 0.0040 0.0040 0.0080 0.010 0.020 0.030 0.030 0.030 0.020 0.020 0.020 0.040 0.040 5 × 10-4 20 × 10-4 46 × 10-4 179 x 10 355 X 10 a If the initial rate data shown are for the rate at which I disappears, -d()/dt, what would be the rate of disappearance of CroOin Trial 1? 1.7 × 104 M/s Determine the experimental rate law for the reaction? Rate = k(Ch072)(1.)(H)2 b) 3A TI-84 Plus CE

Explanation / Answer

Part a

From the stoichiometry of the reaction

Relation between rates

-(Rate of disappearance of Cr2O7(2-)) /1 = -(Rate of disappearance of I-) /6

For trial 1

-(Rate of disappearance of I-) = 5 x 10^-4

-(Rate of disappearance of Cr2O7(2-)) /1 =( 5 x 10^-4) /6

Rate of disappearance of Cr2O7(2-) = 8.33 x 10^-5 M

Part b

A = Cr2O7(2-)

B = I-

C = H+

Let assume the Rate law r = k [A]a [B]b [C]c

From trial 1 and 2

r1/r2 =[A]1a [B]1b [C]1c / [A]2a [B]2b [C]2c

5/20 = (0.010/0.020)b

0.25 = 0.5^b

0.5^2 = 0.5^b

b = 2

From trial 4 and 5

r4/r5 =[A]4a [B]4b [C]4c / [A]5a [B]5b [C]5c

179/355 = (0.0040/0.0080)a

0.50 = 0.5^a

0.5^1 = 0.5^a

a = 1

From trial 3 and 4

r3/r4 =[A]3a [B]3b [C]3c / [A]4a [B]4b [C]4c

46/179 = (0.020/0.040)c

0.25 = 0.5^c

0.5^2 = 0.5^c

c = 2

Rate = r = k [A] [B]2 [C]2

r = k [Cr2O7(2-)] [I-]2 [H+]2

Part C

From trial 1

r1 = k [Cr2O7(2-)]1 [I-]12 [H+]12

5 x 10^-4 = k x (0.0040) x (0.010)2 x (0.020)2

k = 3.125 x 10^6 L4 mol-4 s-1

The given answer of part C is wrong

The correct answer is

k = 3.125 x 10^6 L4 mol-4 s-1

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