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... 84 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... 22 sec
Calculate
The Experimental value of exponent a..._______________ (a)
2) If the exponents a, b, and c have the values:
a = 0; b = 2; c = 1
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...8.3 mL
0.0010 M Sodium thiosulfate...8.3 mL
Distilled water...18.0 mL
0.0400 M Potassium Bromate...6.9 mL
0.100 M Hydrochloric acid...5.7 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 = -4.20
and 1/T = 0.00395, log rate = -6.00.
CALCULATE:
The Slope of the line..._____________ (d)
The Activation Energy (Ea) (Kj/mol)...________________(e)
To answer the given questions, we will use the rate equation provided:
RATE = ([Na2S2O3]/time) = k [I-]^a [BrO3-]^b [HCl]^c
1A) The experimental value of exponent a can be determined by comparing the concentrations of [I-] in the two experiments and the corresponding times to turn blue.
In the first experiment, [I-] is 0.0100 M and the time is 84 sec.
In the second experiment, [I-] is the same, 0.0100 M, but the time is 22 sec.
By comparing the two experiments, we can see that the concentration of [I-] does not affect the rate of the reaction since [I-] is the same in both experiments. Therefore, the exponent a is 0.
1B) The rate constant k can be calculated using the formula:
k = (RATE) / ([I-]^a [BrO3-]^b [HCl]^c)
Let's substitute the given values from experiment 1A:
RATE = 1 / 84 sec
[I-] = 0.0100 M
[BrO3-] = 0.0400 M
[HCl] = 0.100 M
Therefore,
k = (1 / 84 sec) / (0.0100 M^0 * 0.0400 M^2 * 0.100 M^1)
Simplifying the expression,
k = (1 / 84 sec) / (0.000040 M^2)
Calculating the value of k will give us the answer for (b).
3) The time to turn blue can be calculated using the rate equation and rate constant from problem 2.
Using the given volumes and concentrations:
[I-] = 0.0100 M * (8.3 mL / 10.0 mL) = 0.0083 M
[BrO3-] = 0.0400 M * (6.9 mL / 10.0 mL) = 0.0276 M
[HCl] = 0.100 M * (5.7 mL / 10.0 mL) = 0.057 M
Let's substitute these values into the rate equation:
RATE = k [I-]^a [BrO3-]^b [HCl]^c
To calculate the time to turn blue, we need to rearrange the equation to solve for time:
time = [Na2S2O3] / (RATE) = [Na2S2O3] / (k [I-]^a [BrO3-]^b [HCl]^c)
Substituting the given values,
time = (0.00100 M * 10.0 mL) / (k * 0.0083 M^0 * 0.0276 M^2 * 0.057 M^1)
Calculating the value of time will give us the answer for (c).
4) To calculate the slope of the line and the activation energy, we need to use the Arrhenius equation:
ln(k) = - (Ea / R) * (1/T) + ln(A)
Given the two points (1/T, log rate) on the line:
Point 1: (1/T = 0.00300, log rate = -4.20)
Point 2: (1/T = 0.00395, log rate = -6.00)
We can set up two equations using these points:
ln(k1) = - (Ea / R) * (0.00300) + ln(A)
ln(k2) = - (Ea / R) * (0.00395) + ln(A)
Subtracting the second equation from the first equation will eliminate ln(A):
ln(k1/k2) = (Ea / R) * (0.00395 - 0.00300)
We know that k1/k2 = 10^[(log rate1 - log rate2)], so let's substitute:
ln(10^[(log rate1 - log rate2)]) = (Ea / R) * (0.00395 - 0.00300)
Simplifying the equation,
[(log rate1 - log rate2)] = (Ea / R) * (0.00395 - 0.00300)
The slope of the line is given by (Ea / R), so we can rearrange the equation to solve for the slope:
(Ea / R) = [(log rate1 - log rate2)] / (0.00395 - 0.00300)
Calculating the value of the slope will give us the answer for (d).
The activation energy (Ea) can be calculated by multiplying the value of (Ea / R) by the gas constant (R, which is approximately 8.314 J/(mol·K)). Therefore, the answer for (e) can be obtained by multiplying the slope from (d) by the value of R.
I hope this explanation helps you understand how to solve these problems.