Name QUESTIONS 1. Consider this reaction and the experimental quantities given i
ID: 543538 • Letter: N
Question
Name QUESTIONS 1. Consider this reaction and the experimental quantities given in the BrO (ap) Run 000200M 0.00200M 0.0200M HO KBrO, (mL) HC(mL) (mL) volume (mL) Total|Initial Rate-- B (Ms volume (mLy baicia oeBoa 1.44× 10-12 110.0 10.0 10.0 70.0 100.0 2 20. 3 10.0 4 10.0 2.88 × 10-12 10.0 10.0 60.0 100.0 20.0 10.0 .100.0 10.0 20.0 60.0 100.0 2.88× 10-12 5.76 × 10-12 a. Calculate the molarities in each of the reaction mixtures. Run No. KBr KBrO,M HCI b. Determine the order of each reactant. (Show work below.) order of [Br] order of [BrO,1 order of [H']Explanation / Answer
Volume of KBr- 10ml, it molarity = 0.002M, moles of KBr= Molarity* Volume in L= 0.002*10/1000=0.00002. in the 100ml of final solution, no of moles remains the same, but due to dilution to 100ml, the concentration drops down and this concentration = 0.0002
The volume and concentrations of HCl and KBrO3 are same as KBr in runno-1. Hence
Their concentrations remain at 0.0002 M
Calculations are done for other concentrations on similar grounds shown above and tabulated.
Moles = molarity* Volume in Lliters
Final concentration = Moles/ Volume in L= moles/ (100/1000)L= moles/0.1
Run-1
After dilution to 100ml, the molarity of KBr =0.00002/0.1 =0.0002
Moles of KBRo3= 0.0002 after dilution
After dilution to 100ml, molarity of HCl =0.0002
Run-2
Molarity of KBr 20*0.002/1000 =0.0004
Molarity of KBrO3 after dilution = 0.0002 ( same as run-1)
=0.0002 same run-1
Run-3
0.0002 ( same as run-1)
=0.0004
=0.0002
Run-4
=0.0002
=0.0002
=0.0004
Let the rate law for formation of Br2 be represented as r= K[Br-]a [BrO3-]b[ H+]c,
Where K is rate constant and a, b and c are orders of the reaction with respect o [Br-], [BrO3-] and [H+] respectively
From run-1, K[0.0002]a[ 0.0002]b [0.0002]c = 1.44*10-12 (1)
From run-2, K[0.0004]a[ 0.0002]b [0.0002]c = 2.88*10-12 (2)
Eq.2/Eq.1, 2a=2, a= 1
From run-3 K[0.0002]a[ 0.0002]b [0.0002]c = 2.88*10-12 (3)
Eq.3/Eq.1 2b=2, b= 1
From run-4, K[0.0002]a[ 0.0002]b [0.002]c = 5.76*10-12 (3)
Eq.4/Eq.1, 2c= 4, C =2
The rate law becomes, K[ BrO3-][Br-2] [H+]2
From run-1
K[ 0.0002] [0.0002] [0.0002]2= 1.44*10-12
K= 900/M3.sec
The rate law becomes, r= 900[Br-] [BrO3-][H+]2
when [Br-] = [BrO3-] =3*10-3, [H+]= 2*10-3
the rate of formation of Br2= 900*3*10-3*3*10-3*(2*10-3)2=3.24*10-8 M/sec
from the reaction stoichiometry
rBr2/ 3= -rBr-/5
where 3 and 5 are coefficients of Br2 and Br- in the reaction
rBr2 is rate of formation of Br2 and –rbRO3 is rate of decomposition of BrO3
-rBrO3= 5*3.24*10-8/3= 5.4 *10-8 M/s
Run-1
After dilution to 100ml, the molarity of KBr =0.00002/0.1 =0.0002
Moles of KBRo3= 0.0002 after dilution
After dilution to 100ml, molarity of HCl =0.0002
Run-2
Molarity of KBr 20*0.002/1000 =0.0004
Molarity of KBrO3 after dilution = 0.0002 ( same as run-1)
=0.0002 same run-1
Run-3
0.0002 ( same as run-1)
=0.0004
=0.0002
Run-4
=0.0002
=0.0002
=0.0004
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