2. Optional Assume that the reaction studied in Problem 1 isFe*(aq) + 2 SCN"(aq) =Fe(SCN);*(aq). Find K, for this reaction, given the data in Problem 1. a. Formulate the expression for K, for the alternate reaction just cited. b. Find K, as you did in Problem 1; take due account of the fact that two moles SCN¯ are used up per mole Fe(SCN),* formed as you carry out the following calculations: Step 1 Results are as in Problem 1. Step 2 How many moles of Fe(SCN),* are in the mixture at equilibrium? (This will be the same as in Problem 1.) _moles Fe(SCN),* How many moles of Fe* and SCN¯ are used up in making the Fe(SCN),*? moles Fe";. moles SCN- Step 3 How many moles of Fe* and SCN" remain in solution at equilibrium? Use the results of Steps 1 and 2, noting that moles SCN¯ at equilibrium = original moles SCN" – (2 × moles Fe(SCN),*). moles Fe": moles SCN" Step 4 What are the concentrations of Fe*. SCN¯, and Fe(SCN),* at equilibrium? (Use Eq. 3 and the results of Step 3.) [Fe*] = _M; [SCN"] =, M; [Fe(SCN),*] = M Step 5 Calculate K, based on the assumption that the alternate reaction occurs. (Use the answer to Part 2a.) K =
2. Optional Assume that the reaction studied in Problem 1 isFe*(aq) + 2 SCN"(aq) =Fe(SCN);*(aq). Find K, for this reaction, given the data in Problem 1. a. Formulate the expression for K, for the alternate reaction just cited. b. Find K, as you did in Problem 1; take due account of the fact that two moles SCN¯ are used up per mole Fe(SCN),* formed as you carry out the following calculations: Step 1 Results are as in Problem 1. Step 2 How many moles of Fe(SCN),* are in the mixture at equilibrium? (This will be the same as in Problem 1.) _moles Fe(SCN),* How many moles of Fe* and SCN¯ are used up in making the Fe(SCN),*? moles Fe";. moles SCN- Step 3 How many moles of Fe* and SCN" remain in solution at equilibrium? Use the results of Steps 1 and 2, noting that moles SCN¯ at equilibrium = original moles SCN" – (2 × moles Fe(SCN),*). moles Fe": moles SCN" Step 4 What are the concentrations of Fe*. SCN¯, and Fe(SCN),* at equilibrium? (Use Eq. 3 and the results of Step 3.) [Fe*] = _M; [SCN"] =, M; [Fe(SCN),*] = M Step 5 Calculate K, based on the assumption that the alternate reaction occurs. (Use the answer to Part 2a.) K =
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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answer question 2. on the second picture using the data from picture 1.
![Advance Study Assignment: Determination of the Equilibrium Constant
for a Chemical Reaction
1. A student mixes 5.00 mL 2.00 x 10-3 M Fe(NO,), in 1 M HNO, with 3.00 mL 2.00 X 10-M KSCN and
2.00 mL of water. She finds that in the equilibrium mixture the concentration of FeSCN?* is 7.0 x 10-5 M.
Find K, for the reaction Fe+(aq) + SCN"(aq) = Fe(SCN)?*(aq).
Step 1 Find the number of moles Fe* and SCN initially present. (Use Eq. 3.)
moles A: M, V
3
ニ210。5.l6|2.103.3.10°2
< |. 10-5m
-3
Ix10-5
moles Fe+,6x10-
moles SCN-
Step 2 How many moles of FeSCN?* are in the mixture at equilibrium? What is the volume of the equi-
librium mixture? (Use Eq. 3.)
Sml + 3ril +2 mL =10 mL
7.10- •10.10-
10
mL: 7.0X1D
How many moles of Fe* and SCN are used up in making the FeSCN?*?
moles FeSCN2+
7,0x10 moles Fe*:.
7.0x10
moles SCN
Step 3 How many moles of Fe* and SCN- remain in the solution at equilibrium? (Use Eq. 4 and the
results of Steps 1 and 2.)
エn 61o
1105 - 7,0 • 10
6.106-7.0 jot
9.3. 10-6
9.3x106
moles Fe":5.3 XIO moles SCN-
F。うt
Step 4 What are the concehtrations of Fe*, SCN", and FeSCN* at equilibrium? (Use Eq. 3 and the
results of Step 2 and Step 3.)
7x10-7 mol
9.3 x 10-4
-4
.5
[Fe*] = 9.3x10 M: [SCN-] =2:3x1O M: [FESCN*] =
-4
: 5.3x10
Step 5 What is the value of K, for the reaction? (Use Eq. 2 and the results of Step 4.)
M
*7.10-5.
K, = 192o102
- 0,142 .10 3
(continued on following page)
142,0)
こ](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff72e9aba-d7d1-4b51-81b7-9e8888abf351%2F168e9006-6f27-4f58-a90b-55ef7bef42d7%2Ftyirs7a_processed.png&w=3840&q=75)
Transcribed Image Text:Advance Study Assignment: Determination of the Equilibrium Constant
for a Chemical Reaction
1. A student mixes 5.00 mL 2.00 x 10-3 M Fe(NO,), in 1 M HNO, with 3.00 mL 2.00 X 10-M KSCN and
2.00 mL of water. She finds that in the equilibrium mixture the concentration of FeSCN?* is 7.0 x 10-5 M.
Find K, for the reaction Fe+(aq) + SCN"(aq) = Fe(SCN)?*(aq).
Step 1 Find the number of moles Fe* and SCN initially present. (Use Eq. 3.)
moles A: M, V
3
ニ210。5.l6|2.103.3.10°2
< |. 10-5m
-3
Ix10-5
moles Fe+,6x10-
moles SCN-
Step 2 How many moles of FeSCN?* are in the mixture at equilibrium? What is the volume of the equi-
librium mixture? (Use Eq. 3.)
Sml + 3ril +2 mL =10 mL
7.10- •10.10-
10
mL: 7.0X1D
How many moles of Fe* and SCN are used up in making the FeSCN?*?
moles FeSCN2+
7,0x10 moles Fe*:.
7.0x10
moles SCN
Step 3 How many moles of Fe* and SCN- remain in the solution at equilibrium? (Use Eq. 4 and the
results of Steps 1 and 2.)
エn 61o
1105 - 7,0 • 10
6.106-7.0 jot
9.3. 10-6
9.3x106
moles Fe":5.3 XIO moles SCN-
F。うt
Step 4 What are the concehtrations of Fe*, SCN", and FeSCN* at equilibrium? (Use Eq. 3 and the
results of Step 2 and Step 3.)
7x10-7 mol
9.3 x 10-4
-4
.5
[Fe*] = 9.3x10 M: [SCN-] =2:3x1O M: [FESCN*] =
-4
: 5.3x10
Step 5 What is the value of K, for the reaction? (Use Eq. 2 and the results of Step 4.)
M
*7.10-5.
K, = 192o102
- 0,142 .10 3
(continued on following page)
142,0)
こ
![2. Optional Assume that the reaction studied in Problem 1 isFe³*(aq) + 2 SCN"(aq) =Fe(SCN),*(aq).
Find K, for this reaction, given the data in Problem 1.
a. Formulate the expression for K, for the alternate reaction just cited.
K¸ =
b. Find K, as you did in Problem 1; take due account of the fact that two moles SCN¯ are used up per
mole Fe(SCN),* formed as you carry out the following calculations:
Step 1 Results are as in Problem 1.
Step 2 How many moles of Fe(SCN),* are in the mixture at equilibrium? (This will be the same as in
Problem 1.)
moles Fe(SCN),*
How many moles of Fe* and SCN- are used up in making the Fe(SCN), *?
moles Fe*;
moles SCN-
Step 3 How many moles of Fe* and SCN¯ remain in solution at equilibrium? Use the results of Steps 1
and 2, noting that moles SCN" at equilibrium = original moles SCN- (2 × moles Fe(SCN),*).
moles Fe*;
_moles SCN"
Step 4 What are the concentrations of Fe³*. SCN¯, and Fe(SCN),* at equilibrium? (Use Eq. 3 and the
results of Step 3.)
[Fe*] =.
_M; [Fe(SCN),*] = _
M; [SCN"] =
M
Step 5 Calculate K, based on the assumption that the alternate reaction occurs. (Use the answer to Part 2a.)
K =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff72e9aba-d7d1-4b51-81b7-9e8888abf351%2F168e9006-6f27-4f58-a90b-55ef7bef42d7%2Fi7kdwmc_processed.png&w=3840&q=75)
Transcribed Image Text:2. Optional Assume that the reaction studied in Problem 1 isFe³*(aq) + 2 SCN"(aq) =Fe(SCN),*(aq).
Find K, for this reaction, given the data in Problem 1.
a. Formulate the expression for K, for the alternate reaction just cited.
K¸ =
b. Find K, as you did in Problem 1; take due account of the fact that two moles SCN¯ are used up per
mole Fe(SCN),* formed as you carry out the following calculations:
Step 1 Results are as in Problem 1.
Step 2 How many moles of Fe(SCN),* are in the mixture at equilibrium? (This will be the same as in
Problem 1.)
moles Fe(SCN),*
How many moles of Fe* and SCN- are used up in making the Fe(SCN), *?
moles Fe*;
moles SCN-
Step 3 How many moles of Fe* and SCN¯ remain in solution at equilibrium? Use the results of Steps 1
and 2, noting that moles SCN" at equilibrium = original moles SCN- (2 × moles Fe(SCN),*).
moles Fe*;
_moles SCN"
Step 4 What are the concentrations of Fe³*. SCN¯, and Fe(SCN),* at equilibrium? (Use Eq. 3 and the
results of Step 3.)
[Fe*] =.
_M; [Fe(SCN),*] = _
M; [SCN"] =
M
Step 5 Calculate K, based on the assumption that the alternate reaction occurs. (Use the answer to Part 2a.)
K =
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