Question

In: Chemistry

RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c 1A) You mix together in the proper...

RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
1A) You mix together in the proper manner the following:
10.0 mL of .0100 M Potassium Iodide
10.0 mL of .00100 M Sodium thiosulfate
10.0 mL of water
10.0 mL of .0400 M Potassium Bromate
10.0 mL of .100 M HCl
The time to turn blue is... 53 sec
1B) You also mix together in the proper manner the following:
10.0 mL of .0100 M Potassium Iodide
10.0 mL of .00100 M Sodium thiosulfate
10.0 mL of .0400 M Potassium Bromate
20.0 mL of .100 M HCl
The time to turn blue is... 14 sec
Calculate:
The Experimental value of exponent c..._______________ (a)

2) If the exponents a, b, and c have the values:
a = 1; b = 0; c = 2
CALCULATE:
Rate constant k for data in 1A...______________ (b)

3) You mix together in the proper manner the following VOLUMES:
0.0100 M Potassium iodide...14.7 mL
0.0010 M Sodium thiosulfate...5.6 mL
Distilled water...22.7 mL
0.0400 M Potassium Bromate...11.7 mL
0.100 M Hydrochloric acid...5.6 mL
Using the rate constant and rate equation from problem 2),

CALCULATE the time to turn blue...__________________ sec (c)

4) A plot of log Rate vs 1/T gives a best straight line which goes through....
   1/T = 0.00300, log rate = -4.46
and 1/T = 0.00352, log rate = -6.00.
CALCULATE:
The Slope of the line..._____________ (d)

The Activation Energy (Ea) (Kj/mol)...________________(e)

Solutions

Expert Solution

1. the concentration of all the reactants are kept constant in second experiment except HCl.

The concentration of HCl was made double which reduces the time approx four times to the first experiment. ( 53/14 = 3.78)

So the rate increases 3.78 times that of first experiment

Rate1 /Rate2 = {[HI]1 /[ HI2]}c = 1/3.78 = [0.1 X 10 / 0.1 X 20 ]c

3.78 = (2)c

log 3.78 = c log 2

0.577 = c X 0.3010

c = 1.91

The Experimental value of exponent c = 1.91

b. RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c

1. 10.0 mL of .0100 M Potassium Iodide, so moles = molarity X volume = 0.01 X 10 /1000 =

or concenration = moles / total volume =   0.0001    X1000 / 50 mL =0.002 M
2. 10.0 mL of .00100 M Sodium thiosulfate; so moles = molarity X volume = 0.01 X 10 /1000 =

or concenration = moles / total volume =   0.00001  X1000 / 50 mL =0.0002M
3.10.0 mL of water
4. 10.0 mL of .0400 M Potassium Bromate ; so moles = molarity X volume = 0.04 X 10 /1000 = 0.0004

or concenration = moles / total volume =    0.0004   X1000 / 50 mL =0.008
5. 10.0 mL of .100 M HCl ; so moles = molarity X volume = 0.1 X 10 /1000 = 0.001

or concenration = moles / total volume =  0.001     X1000 / 50 mL =0.02M

RATE = ([ Na2S2O3]/time) = k [I-] [BrO3-]0 [HCl]2 = 0.0002 / 53 = K X 0.0002 X (0.02)2

3.77 X 10^-6 = K X8X10^-8

K = 47.1 moles ^-2 X L-1 X second^-1

................................................

c. RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c

total volume =14.7 +5.6+22.7+11.7 + 5.6 = 60.3

i) 0.0100 M Potassium iodide...14.7 mL

so moles = molarity X volume = 0.01 X 14.7 /1000 =.147 / 1000

or concentration = moles / total volume =   0.000147X1000 / 60.3=0.0024 M

ii) 0.0010 M Sodium thiosulfate...5.6 mL

so moles = molarity X volume = 0.001 X 5.6 /1000 =

or concentration = moles / total volume =   0.0056   / 60.3 mL = 9.2 X 0^-5M

iii) Distilled water...22.7 mL
iv) 0.0400 M Potassium Bromate...11.7 mL

so moles = molarity X volume = 0.04 X 11.7 /1000 = 0.468 /1000

or concentration = moles / total volume =   0.468 / 60.3 mL =0.0077 M

v) 0.100 M Hydrochloric acid...5.6 mL

so moles = molarity X volume = 0.1 X 5.6 /1000 = 0.56 / 1000

or concentration = moles / total volume =   0.56    / 60.3 mL =0.0092 M

Rate constant = K = 47.1 ( from part b)

RATE = ([ Na2S2O3]/time) = k [I-] [BrO3-]0 [HCl]2 = 9.2 X 10-5/ time = 47.1 X 0.0024 X (0.0092)2

= 9.56 X 10^-6= 9.2 X 10-5/ time

Time = 9.6 seconds

.......................

d) For the given grap ( which is Arrehenius graph)

Slope =lnK2- lnK1 / 1/T2 - 1/T1 = (-6) - ( -4.46) / (0.00352 - 0.003) = -1.54 / 0.00052= -2961.53

Now , Slope = = -Ea / R

Ea = activation energy

Ea = 2961.53 X 8.314 =24622.16 J / mole = 24.622 KJ / mole


Related Solutions

If the rate law for the clock reaction is: Rate = k [ I-] [ BrO3...
If the rate law for the clock reaction is: Rate = k [ I-] [ BrO3 -] [H+] A clock reaction is run with the following initial concentrations: [I-] [BrO3-] [H+] [S2O32-] 0.002 0.008 0.02 0.0001 The reaction time is 28 seconds Calculate k in the rate law: Also: Rate = k [ I-] [ BrO3 -] [H+] A clock reaction is run at 19 ºC with the following initial concentrations [I-] [BrO3-] [H+] [S2O32-] 0.002 0.008 0.02 0.0001 Then...
I. Answer part A, B, and C 1a) Return true if the given int parameter is...
I. Answer part A, B, and C 1a) Return true if the given int parameter is both positive and less than 10, and false otherwise. Remember that 0 is not positive! positiveLessThan10(10) → false positiveLessThan10(9) → true positiveLessThan10(11) → false 1b) Return true if the parameters are in ascending order, from left to right. Two values that are equal are considered to be in ascending order. Remember, you can just type return true or return false to return true or...
How many proper subsets are there for this set {A,B,C,D,E,F,G,H,I}?
How many proper subsets are there for this set {A,B,C,D,E,F,G,H,I}?
For a reaction A +B → C, the experimental rate law is found to be R=k[A]1[B]1/2. Find the rate of the reaction when [A] = 0.5 M, [B] = 0.1 M and k=0.03
For a reaction A +B → C, the experimental rate law is found to be R=k[A]1[B]1/2. Find the rate of the reaction when [A] = 0.5 M, [B] = 0.1 M and k=0.03.
The reaction 2A + B → C obeys the rate law –rB = kCB2, where k...
The reaction 2A + B → C obeys the rate law –rB = kCB2, where k = 0.25 L/mol.s at 75˚C. A is available as a 1.5 M solution and B as a 1 M solution. a) A 26 L/s stream of solution A and 30 L/s stream of solution B are combined just before being introduced into a CSTR. If a 150 L reactor is available, how many mole/h of C could be produced? b) You receive some financing...
Consider the following reaction at 283 K: 2A + B → C + D where rate...
Consider the following reaction at 283 K: 2A + B → C + D where rate = k[A][B]2. An experiment was performed where [A]o = 2.67 M and [B]o = 0.00241 M. A plot of 1/[B] vs. time has a slope of 10.01. What will the rate of this reaction be when [A] = [B] = 0.345 M?
For the reaction A +B -->2C, suppose that you find that the rate=k[A][B]^2. Indicate how the...
For the reaction A +B -->2C, suppose that you find that the rate=k[A][B]^2. Indicate how the overall reaction rate will change if you... double the concentration of A double the concentration of B double the concentration of both A and B double the concentration of A and halve the concentration of B
i just need the national saving rate for the last two, b and c, I answered...
i just need the national saving rate for the last two, b and c, I answered everything else by myself and i just need the national saving for B and C In each part that follows, use the economic data given to find national saving, private saving, public saving, and the national saving rate. a. Household saving = 200     Business saving = 400      Government purchases of goods and services = 100      Government transfers and interest payments = 100      Tax collections...
I need the answer for PART B and PART C of this question: You have recently...
I need the answer for PART B and PART C of this question: You have recently been appointed management accountant for Rugby Coffee Mugs Pty Ltd. The company commenced its operations on 1 July 2019 manufacturing one size coffee mugs with individual club names and club logos of rugby union clubs playing in the New South Wales, Queensland, Victoria and Western Australia local rugby union competition. The company currently does not have any management accounting controls and part of your...
you can see the answer a, b, and c. I have a question about d, e,...
you can see the answer a, b, and c. I have a question about d, e, and f only how to solve theses questions Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and work-piece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. The article "Variables Affecting Mist Generation from Metal Removal Fluids" (Lubrication Engr., 2002: 10-17) gave the accompanying data...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT