It is given that for a common type of reaction the value of K < < 1 . The assumption made when solving the equilibrium concentrations for reactions with small K values and the procedure followed if the ‘ 5 % rule’ fails are to be stated. Concept introduction: If the error in the equilibrium concentration calculated is less than 5 % , the assumption made is stated to be correct. This is known as the ‘ 5 % rule’. If the error exceeds the 5 % value, then the quadratic formula is used for calculating the equilibrium concentrations. To determine: The assumption made when solving the equilibrium concentrations for reactions with small K values and the procedure followed if the ‘ 5 % rule’ fails.
It is given that for a common type of reaction the value of K < < 1 . The assumption made when solving the equilibrium concentrations for reactions with small K values and the procedure followed if the ‘ 5 % rule’ fails are to be stated. Concept introduction: If the error in the equilibrium concentration calculated is less than 5 % , the assumption made is stated to be correct. This is known as the ‘ 5 % rule’. If the error exceeds the 5 % value, then the quadratic formula is used for calculating the equilibrium concentrations. To determine: The assumption made when solving the equilibrium concentrations for reactions with small K values and the procedure followed if the ‘ 5 % rule’ fails.
Solution Summary: The author explains how the quadratic equation method is used for solving the equilibrium concentrations for reactions with small K values and the procedure followed if the 5% rule fails.
Interpretation: It is given that for a common type of reaction the value of
K<<1. The assumption made when solving the equilibrium concentrations for reactions with small
K values and the procedure followed if the ‘
5% rule’ fails are to be stated.
Concept introduction: If the error in the equilibrium concentration calculated is less than
5%, the assumption made is stated to be correct. This is known as the ‘
5% rule’. If the error exceeds the
5% value, then the quadratic formula is used for calculating the equilibrium concentrations.
To determine: The assumption made when solving the equilibrium concentrations for reactions with small
K values and the procedure followed if the ‘
5% rule’ fails.
Instructions: Complete the questions in the space provided. Show all your work
1. You are trying to determine the rate law expression for a reaction that you are completing at 25°C. You measure
the initial reaction rate and the starting concentrations of the reactions for 4 trials.
BrO³¯ (aq) + 5Br¯ (aq) + 6H* (aq) → 3Br₂ (l) + 3H2O (l)
Initial rate
Trial
[BrO3]
[H*]
[Br]
(mol/L)
(mol/L) | (mol/L)
(mol/L.s)
1
0.10
0.10
0.10
8.0
2
0.20
0.10
0.10
16
3
0.10
0.20
0.10
16
4
0.10
0.10
0.20
32
a.
Based on the above data what is the rate law expression?
b. Solve for the value of k (make sure to include proper units)
2. The proposed reaction mechanism is as follows:
i.
ii.
BrО¸¯ (aq) + H+ (aq) → HBrO3 (aq)
HBrO³ (aq) + H* (aq) → H₂BrO3* (aq)
iii.
H₂BrO³* (aq) + Br¯ (aq) → Br₂O₂ (aq) + H2O (l)
[Fast]
[Medium]
[Slow]
iv.
Br₂O₂ (aq) + 4H*(aq) + 4Br(aq) → 3Br₂ (l) + H2O (l)
[Fast]
Evaluate the validity of this proposed reaction. Justify your answer.
е.
Д
CH3
D*, D20
C.
NaOMe,
Br
Br
Chapter 12 Solutions
Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.