A common type of reaction we will study is that having a very small K value ( K << 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K << 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the “5% rule” is used to check the validity of assuming x (or 2 x , 3 x, and so on) is very small compared to some number. When x (or 2 x . 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
A common type of reaction we will study is that having a very small K value ( K << 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K << 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the “5% rule” is used to check the validity of assuming x (or 2 x , 3 x, and so on) is very small compared to some number. When x (or 2 x . 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
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.
A common type of reaction we will study is that having a very small K value (K << 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K << 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the “5% rule” is used to check the validity of assuming x (or 2 x, 3x, and so on) is very small compared to some number. When x (or 2 x. 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
Using the conditions of spontaneity to deduce the signs of AH and AS
Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy
AS.
Note: if you have not been given enough information to decide a sign, select the "unknown" option.
reaction
observations
conclusions
A
This reaction is always spontaneous, but
proceeds slower at temperatures above
120. °C.
ΔΗ is
(pick one)
AS is
(pick one)
ΔΗ is
(pick one)
B
This reaction is spontaneous except above
117. °C.
AS is
(pick one)
ΔΗ is
(pick one)
This reaction is slower below 20. °C than
C
above.
AS is
|(pick one)
?
18
Ar
1
Calculating the pH at equivalence of a titration
Try Again
Your answer is incorrect.
0/5
a
A chemist titrates 70.0 mL of a 0.7089 M hydrocyanic acid (HCN) solution with 0.4574M KOH solution at 25 °C. Calculate the pH at equivalence. The pK of
hydrocyanic acid is 9.21.
Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added.
pH
=
11.43]
G
00.
18
Ar
B•
Biological Macromolecules
Naming and drawing the products of aldose oxidation and reduction
aw a Fischer projection of the molecule that would produce L-ribonic acid if it were subjected to mildly oxidizing reaction conditions.
Click and drag to start drawing a
structure.
X
AP
‡
1/5
Naor
Explanation
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