Interpretation Highly toxic disulfur decafluoride decomposes by a free radical process: S 2 F 10 (g) ⇌ SF 4 (g) + SF 6 (g) . In a study of the decomposition, S 2 F 10 was placed in a 2.0 L flask and heated to 100 o C ; [S 2 F 10 ] was 0.50 M at equilibrium. More S 2 F 10 was added, and when equilibrium was retained, the concentration of S 2 F 10 was 2.5 M. How the concentrations change of [SF 4 ] and [SF 6 ] affects the equilibrium has to be determined. Concept Introduction Law of Chemical Equilibrium The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients. Equilibrium Constant Consider a reaction, aA+bB ⇔ cC+dD Forward reaction rate K f = [A] a [B] B Backward reaction rate K b = [C] c [D] d At equilibrium, the rate of forward reaction = rate of backward reaction K eq = K f K b K f K b = [C] c [D] d [A] a [B] B K eq = K f K b = [C] c [D] d [A] a [B] B K eq is the equilibrium constant.
Interpretation Highly toxic disulfur decafluoride decomposes by a free radical process: S 2 F 10 (g) ⇌ SF 4 (g) + SF 6 (g) . In a study of the decomposition, S 2 F 10 was placed in a 2.0 L flask and heated to 100 o C ; [S 2 F 10 ] was 0.50 M at equilibrium. More S 2 F 10 was added, and when equilibrium was retained, the concentration of S 2 F 10 was 2.5 M. How the concentrations change of [SF 4 ] and [SF 6 ] affects the equilibrium has to be determined. Concept Introduction Law of Chemical Equilibrium The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients. Equilibrium Constant Consider a reaction, aA+bB ⇔ cC+dD Forward reaction rate K f = [A] a [B] B Backward reaction rate K b = [C] c [D] d At equilibrium, the rate of forward reaction = rate of backward reaction K eq = K f K b K f K b = [C] c [D] d [A] a [B] B K eq = K f K b = [C] c [D] d [A] a [B] B K eq is the equilibrium constant.
Highly toxic disulfur decafluoride decomposes by a free radical process: S2F10(g) ⇌ SF4(g) + SF6(g). In a study of the decomposition, S2F10 was placed in a 2.0 L flask and heated to 100oC; [S2F10] was 0.50 M at equilibrium. More S2F10 was added, and when equilibrium was retained, the concentration of S2F10 was 2.5 M. How the concentrations change of [SF4] and [SF6] affects the equilibrium has to be determined.
Concept Introduction
Law of Chemical Equilibrium
The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.
Equilibrium Constant
Consider a reaction,
aA+bB⇔cC+dD
Forward reaction rate Kf= [A]a[B]B
Backward reaction rate Kb= [C]c[D]d
At equilibrium, the rate of forward reaction = rate of backward reaction
Draw the curved-arrow mechanism with the drawings of the molecules, not
just abbreviations.
-NO₂
Sn, HCl (aq)
E
D
H
(CH3CO)₂O
-NH2
CH3
What is/are the product(s) of the following reaction? Select all that apply.
* HI
A
B
C
OD
OH
A
B
OH
D
C
In the image, the light blue sphere represents a mole of hydrogen atoms, the purple or teal spheres represent a mole of a conjugate base. A light blue sphere by itself is H+.
Assuming there is 2.00 L of solution, answer the following:
The Ka of the left & right solution is?
The pH of the left & right solution is?
The acid on the left & right is what kind of acid?
Chapter 17 Solutions
Chemistry: The Molecular Nature of Matter and Change
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