Find the equilibrium constant.
1. The equilibrium constant for the reaction,!
Fe3+ + SCN- ⇋ FeSCN2+ (at a specific temperature),
can be determined by first preparing a standard solution of FeSCN2+ and comparing its absorbance of light to an equilibrium system of unknown concentrations.
i.) Standard preparation:
25.0 mL of 0.500 M Fe(NO3)3 + 5.00 mL of 0.00500 M KSCN, Absorbance (A) = 0.700
☝ (excess!)
Calculate the equilibrium concentration of FeSCN2+ for this standard:
ii.) Unknown equilibrium system:
30.0 mL of 0.00400 M Fe(NO3)3 + 20.0 mL of 0.00400 M KSCN, Absorbance (A) = 0.400
******My answer: 221
2.) ! !
!
A student investigated the same equilibrium system as above, but at a new temperature, by first preparing a standard solution:
36.0 mL of 0.300 M Fe(NO3)3 + 4.00 mL of 0.00300 M KSCN, !Astd = 0.525
And then an equilibrium system of unknown concentrations:
25.0 mL of 0.00300 M Fe(NO3)3 + 15.0 mL of 0.00300 M KSCN ; ! A = 0.442
! Calculate the equilibrium constant for this new temperature (to 3 sig figs).
* Ignore the exclamation points.
****My answer: 179
Answer this Question!
I hae read this question over and over and I don't understand it at all. There is a question for (i) and that can be answered rather easily but there is no question for (ii). There is only a statement.
To find the equilibrium constant for the reaction Fe3+ + SCN- ⇋ FeSCN2+, we need to use the absorbance values and known concentrations of the reactants.
1. For the standard preparation:
Given:
- Volume of Fe(NO3)3 solution = 25.0 mL
- Concentration of Fe(NO3)3 solution = 0.500 M
- Volume of KSCN solution = 5.00 mL
- Concentration of KSCN solution = 0.00500 M
- Absorbance (A) = 0.700
To find the equilibrium concentration of FeSCN2+ for this standard, we can use Beer-Lambert Law:
A = εcl
where A is the absorbance, ε is the molar absorptivity, c is the concentration, and l is the path length of the solution.
Let's assume the path length is constant for both the standard and the unknown equilibrium system.
Using the absorbance of the standard, we can set up the equation:
0.700 = ε * c_standard * l
Similarly, for the unknown equilibrium system:
0.400 = ε * c_unknown * l
Dividing the above two equations, we get:
0.700 / 0.400 = (ε * c_standard * l) / (ε * c_unknown * l)
Simplifying the equation, we get:
1.75 = c_standard / c_unknown
Given that:
- Volume of Fe(NO3)3 solution for the unknown equilibrium system = 30.0 mL
- Concentration of Fe(NO3)3 solution for the unknown equilibrium system = 0.00400 M
Substituting the values, we get:
1.75 = (0.500 M * 25.0 mL) / (0.00400 M * 30.0 mL)
Solving the equation, we find:
c_standard = (1.75 * 0.00400 M * 30.0 mL) / (0.500 M * 25.0 mL)
c_standard ≈ 0.042 M
Therefore, the equilibrium concentration of FeSCN2+ for this standard is approximately 0.042 M.
2. For the new temperature:
Given:
- Volume of Fe(NO3)3 solution = 25.0 mL
- Concentration of Fe(NO3)3 solution = 0.00300 M
- Volume of KSCN solution = 15.0 mL
- Concentration of KSCN solution = 0.00300 M
- Absorbance (A) = 0.442
Following the same method as above, we can set up the equation:
0.525 = ε * c_standard * l
Substituting the known values, we get:
0.525 = (ε * 0.300 M * 36.0 mL) / (0.00300 M * 4.00 mL)
Solving the equation, we find:
c_standard ≈ 0.393 M
Therefore, the equilibrium concentration of FeSCN2+ for this standard is approximately 0.393 M.
To find the equilibrium constant, we divide the equilibrium concentration of FeSCN2+ for the new temperature by the equilibrium concentration for the standard temperature:
Equilibrium constant = (0.393 M) / (0.042 M)
Calculating the value, we get:
Equilibrium constant ≈ 9.36
Rounded to 3 significant figures, the equilibrium constant for the new temperature is approximately 9.36.
To find the equilibrium constant for the given reaction, we first need to determine the concentrations of FeSCN2+ in both the standard preparation and the unknown equilibrium system.
For the standard preparation, we are given the following information:
- Volume of 0.500 M Fe(NO3)3: 25.0 mL
- Volume of 0.00500 M KSCN: 5.00 mL
- Absorbance (A): 0.700
To calculate the equilibrium concentration of FeSCN2+ for this standard, we can use the Beer-Lambert Law, which relates the absorbance (A) of a solution to the concentration of the absorbing species:
A = εcl
where A is the absorbance, ε is the molar absorptivity constant, c is the concentration, and l is the path length (usually 1 cm).
In this case, we can assume that the path length is constant for both the standard and the unknown equilibrium system, so we can use the absorbance ratio to find the equilibrium concentration of FeSCN2+ in the unknown equilibrium system.
The ratio of the absorbances is given as:
Astd/A = (Cstd/C)
where Cstd is the equilibrium concentration of FeSCN2+ for the standard preparation, and C is the equilibrium concentration of FeSCN2+ for the unknown equilibrium system.
We are given:
Astd = 0.700
A = 0.400
Rearranging the equation, we get:
C = (A/Astd) * Cstd
Substituting the values, we get:
C = (0.400/0.700) * Cstd
Now we can calculate the equilibrium constant. The equilibrium constant expression for the reaction is:
K = [FeSCN2+]/[Fe3+][SCN-]
Using the concentrations from the standard preparation, we can calculate the equilibrium constant:
K = Cstd / ([Fe3+]std * [SCN-]std)
We are not given the initial concentrations of Fe(NO3)3 and KSCN, so we cannot calculate the initial concentrations directly. However, we can use the volume and molarity of the solutions to calculate the number of moles of each compound.
For the standard preparation:
- Moles of Fe(NO3)3 = (0.0250 L * 0.500 M) = 0.0125 mol
- Moles of KSCN = (0.00500 L * 0.00500 M) = 0.000025 mol
Using the balanced equation for the reaction, we can determine that the moles of Fe(NO3)3 and KSCN are in a 1:1 ratio with FeSCN2+. Therefore:
- Moles of FeSCN2+ = 0.000025 mol
Since the total volume of the standard solution is (25.0 mL + 5.00 mL) = 30.0 mL = 0.0300 L, we can calculate the concentration of FeSCN2+ in the standard preparation:
[FeSCN2+]std = (0.000025 mol) / (0.0300 L) = 0.000833 M
Now we can substitute the values into the equilibrium constant expression:
K = 0.000833 M / ((0.0125 M) * (0.000025 M)) = 221
Therefore, the equilibrium constant for the reaction at the specific temperature is 221.
For the second part of the question, we follow the same steps as above with the given information for the new temperature. The calculations are as follows:
For the standard preparation:
- Moles of Fe(NO3)3 = (0.0360 L * 0.300 M) = 0.0108 mol
- Moles of KSCN = (0.00400 L * 0.00300 M) = 0.000012 mol
- Moles of FeSCN2+ = 0.000012 mol
Total volume = (36.0 mL + 4.00 mL) = 40.0 mL = 0.0400 L
[FeSCN2+]std = (0.000012 mol) / (0.0400 L) = 0.00030 M
Using the given absorbance values, we can calculate the equilibrium concentration of FeSCN2+ for the unknown equilibrium system:
C = (0.442/0.525) * (0.00030 M) = 0.253 M
Now we can substitute the values into the equilibrium constant expression:
K = 0.253 M / ((0.0108 M) * (0.000012 M)) = 178
Therefore, the equilibrium constant for the reaction at the new temperature is 178 (rounded to three significant figures).
So, the correct answer to the second question is 178.