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, we need to understand the given rate equation and the principles of reaction rates.
The given rate equation is:
RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
Here's how to approach each question:
1A) Given the following mixture:
- 10.0 mL of 0.0100 M Potassium Iodide
- 10.0 mL of 0.00100 M Sodium thiosulfate
- 10.0 mL of water
- 10.0 mL of 0.0400 M Potassium Bromate
- 10.0 mL of 0.100 M HCl
The time to turn blue is given as 61 seconds.
To find the experimental value of exponent b:
We can use the given rate equation and rearrange it as follows:
RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
Since the concentration of Na2S2O3 is not provided, it can be assumed to be constant. So, we have:
RATE ∝ [I-]a [BrO3-]b [HCl]c
By comparing the two data sets given, 1A and 1B, we can determine the effect of changing the concentration of [BrO3-]. From 1A to 1B, the concentration of [BrO3-] increases, resulting in a shorter time to turn blue. This suggests that the concentration of [BrO3-] has an inverse relationship with the rate (time to turn blue).
To calculate the experimental value of exponent b:
Since [BrO3-] is the only changing factor between 1A and 1B, we can divide the rate (61 sec in 1A) by the rate (14 sec in 1B) and take the ratio:
61 sec / 14 sec = ([BrO3-]1A / [BrO3-]1B )^b
Simplifying the equation further, we have:
(61 / 14)^(1/b) = ([BrO3-]1A / [BrO3-]1B )
Take the logarithm of both sides and solve for b using:
log(61 / 14) = log([BrO3-]1A / [BrO3-]1B / b)
Now, plug in the values, calculate the logarithm, and solve for b.
2) If the exponents a, b, and c have the values: a = 2; b = 2; c = 0
We are given the values of the exponents a, b, and c. To calculate the rate constant k, we need to use the rate equation and the values of the exponents:
Rate = k [I-]^a [BrO3-]^b [HCl]^c
Substituting the given values:
Rate = k [I-]^2 [BrO3-]^2 [HCl]^0
Since [HCl]^0 is equal to 1, it can be ignored:
Rate = k [I-]^2 [BrO3-]^2
Now, we can use the known rate value from 1B and the calculated b to solve for k:
Rate1B = k [I-]^2 [BrO3-]^2
3) Given the following mixture:
- 9.2 mL of 0.0100 M Potassium iodide
- 6.5 mL of 0.0010 M Sodium thiosulfate
- 16.8 mL of distilled water
- 13.9 mL of 0.0400 M Potassium Bromate
- 6.1 mL of 0.100 M Hydrochloric acid
Using the rate constant (k) and rate equation values from problem 2), we can calculate the time to turn blue:
Rate = k [I-]^2 [BrO3-]^2
Plug in the given concentrations ([I-], [BrO3-]) and solve for time using the rate equation and the rate constant obtained in problem 2).
These steps should help you solve the given problems. Make sure to perform the necessary calculations with appropriate units and use the correct significant figures throughout the process.