Given the rate equation...
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... 61 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
20.0 mL of .0400 M Potassium Bromate
10.0 mL of .100 M HCl
The time to turn blue is... 14 sec
Calculate:
The Experimental value of exponent b..._______________ (a)
2) If the exponents a, b, and c have the values:
a = 2; b = 2; c = 0
CALCULATE:
Rate constant k for data in 1B...______________ (b)
3) You mix together in the proper manner the following VOLUMES:
0.0100 M Potassium iodide...9.2 mL
0.0010 M Sodium thiosulfate...6.5 mL
Distilled water...16.8 mL
0.0400 M Potassium Bromate...13.9 mL
0.100 M Hydrochloric acid...6.1 mL
Using the rate constant and rate equation from problem 2),
CALCULATE the time to turn blue...__________________ sec (c)
All i need is explanation on how to do and where to plug in the equation because i am so lost.
thank you!
To solve these problems, you'll need to use the rate equation provided: RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c, where "a," "b," and "c" are the exponents of the concentrations of the reactants.
Let's go through each problem step by step:
1A) You mix together the given solutions in the proper manner and observe that the time to turn blue is 61 seconds. You need to use this information to determine the experimental value of exponent "b."
Given concentrations:
[I-] = 0.0100 M
[Na2S2O3] = 0.00100 M
[BrO3-], [HCl], [water] = unknown
Since you are given the rate equation and the time (rate) for this particular reaction, you can solve for the unknown exponent "b" using the known concentrations.
RATE = ([Na2S2O3] / time) = k [I-]a [BrO3-]b [HCl]c
You can rewrite this equation as:
b = (log R2 - log R1) / (log [BrO3-]2 - log [BrO3-]1)
Plug in the values from the given information:
R1 = 61 sec (time for the first solution)
R2 = 14 sec (time for the second solution)
[BrO3-]1 = 0.0400 M (concentration of Potassium Bromate in the first solution)
[BrO3-]2 = ?
b = ?
To solve for b, you need to calculate [BrO3-]2:
[BrO3-]2 = ([Na2S2O3] / R2) * ([I-]a / [Na2S2O3]) * ([BrO3-]1^b / [I-]^a) * ([HCl]^c / [BrO3-]^b)
Since most of the concentrations are not given, you will need to use the initial known concentration ([BrO3-]1 = 0.0400 M) and solve for [BrO3-]2 using the ratio of times and the known concentrations.
Finally, calculate b using the equation mentioned earlier:
b = (log R2 - log R1) / (log [BrO3-]2 - log [BrO3-]1)
This will give you the experimental value of exponent "b."
2) Given that a = 2, b = 2, and c = 0, you need to calculate the rate constant k for the data in 1B.
Using the given concentrations and exponents, plug them into the rate equation and solve for k using the data from 1B:
Rate = k [I-]^2 [BrO3-]^2 [HCl]^0
Since [HCl] has a exponent of 0, it doesn't affect the rate, so you can ignore it.
Rearrange the equation to solve for k:
k = Rate / ([I-]^2 * [BrO3-]^2)
Substitute the given values and the rate observed from 1B into the equation to find k.
3) Given the volumes instead of concentrations, you will need to convert the volumes into concentrations using the known molarity of each solution.
Use the given concentrations and the known rate constant from problem 2 to calculate the time to turn blue for the new set of volumes.
Plug in the known values for concentrations and volumes into the rate equation and solve for time:
Rate = k [I-]^2 [BrO3-]^2
Rearrange the equation to solve for time:
time = ([I-]^2 * [BrO3-]^2) / k
Substitute the given concentrations and volumes into the equation and solve for time.
Remember to pay attention to units and convert volumes to concentrations as needed.
I hope this explanation helps you understand how to approach and solve these problems step by step.