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176 Chemistry: Matter and Change
Solving Problems: A Chemistry Handbook
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 17
Multiply both sides of the equation by [H 2 ][I 2 ].
[HI] 2 ϭ K eq ϫ [H 2 ][I 2 ]
Substitute the known quantities into the equation and solve for [HI].
[HI] 2 ϭ 66.9 ϫ (0.0295)(0.0174) ϭ 0.03434
An extra digit is retained here for accuracy, but the final answer will
be rounded to three digits.
[HI] ϭ ͙0.0343 ෆ4 ෆ ϭ 0.185 mol/L
The equilibrium concentration of HI is 0.185 mol/L.
Practice Problems
10. At a certain temperature, K eq ϭ 0.118 for the following
reaction.
2CH 4 (g) 3 C 2 H 2 (g) ϩ 3H 2 (g)
Calculate these concentrations.
a. [CH 4 ] in an equilibrium mixture with [C 2 H 2 ]
ϭ 0.0812 mol/L and [H 2 ] ϭ 0.373 mol/L
b. [C 2 H 2 ] in an equilibrium mixture with [CH 4 ] ϭ 0.726 mol/L
and [H 2 ] ϭ 0.504 mol/L
c. [H 2 ] in an equilibrium mixture with [CH 4 ] ϭ 0.0492 mol/L
and [C 2 H 2 ] ϭ 0.0755 mol/L
11. A chemist studying the equilibrium N 2 O 4 (g) 3 2NO 2 (g) controls the temperature so that K eq ϭ 0.028. At one equilibrium
position, the concentration of N 2 O 4 is 1.5 times greater than the
concentration of NO 2 . Find the concentrations of the two gases
in mol/L. (Hint: Let x ϭ [NO 2 ] and 1.5x ϭ [N 2 O 4 ] in the
equilibrium constant expression.)
Solubility equilibria The solubility product constant (K sp )
is an equilibrium constant for the dissolving of a sparingly soluble
ionic compound in water. The solubility product constant expression
is the product of the concentrations of the ions with each concentration raised to a power equal to the coefficient of the ion in the
chemical equation. For example, copper(II) hydroxide dissolves in
water according to this equation.
▲
176 Chemistry: Matter and Change
Solving Problems: A Chemistry Handbook
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 17
Multiply both sides of the equation by [H 2 ][I 2 ].
[HI] 2 ϭ K eq ϫ [H 2 ][I 2 ]
Substitute the known quantities into the equation and solve for [HI].
[HI] 2 ϭ 66.9 ϫ (0.0295)(0.0174) ϭ 0.03434
An extra digit is retained here for accuracy, but the final answer will
be rounded to three digits.
[HI] ϭ ͙0.0343 ෆ4 ෆ ϭ 0.185 mol/L
The equilibrium concentration of HI is 0.185 mol/L.
Practice Problems
10. At a certain temperature, K eq ϭ 0.118 for the following
reaction.
2CH 4 (g) 3 C 2 H 2 (g) ϩ 3H 2 (g)
Calculate these concentrations.
a. [CH 4 ] in an equilibrium mixture with [C 2 H 2 ]
ϭ 0.0812 mol/L and [H 2 ] ϭ 0.373 mol/L
b. [C 2 H 2 ] in an equilibrium mixture with [CH 4 ] ϭ 0.726 mol/L
and [H 2 ] ϭ 0.504 mol/L
c. [H 2 ] in an equilibrium mixture with [CH 4 ] ϭ 0.0492 mol/L
and [C 2 H 2 ] ϭ 0.0755 mol/L
11. A chemist studying the equilibrium N 2 O 4 (g) 3 2NO 2 (g) controls the temperature so that K eq ϭ 0.028. At one equilibrium
position, the concentration of N 2 O 4 is 1.5 times greater than the
concentration of NO 2 . Find the concentrations of the two gases
in mol/L. (Hint: Let x ϭ [NO 2 ] and 1.5x ϭ [N 2 O 4 ] in the
equilibrium constant expression.)
Solubility equilibria The solubility product constant (K sp )
is an equilibrium constant for the dissolving of a sparingly soluble
ionic compound in water. The solubility product constant expression
is the product of the concentrations of the ions with each concentration raised to a power equal to the coefficient of the ion in the
chemical equation. For example, copper(II) hydroxide dissolves in
water according to this equation.
▲