General Chemistry Principles, Patterns, and Applications, 2011

Equilibrium constant expression (law of mass action)
Equation 15.8 : K = [ C]c[ D]d[ A]a[ B]b
Equilibrium constant expression for reactions involving gases using partial pressures
Equation 15.17 : Kp = (PC)c(PD)d(PA)a(PB)b
Relationship between K p and K
Equation 15.19: K p= K(RT) Δ n
C O N C E PTUAL P R OBLEMS
1. For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the
equilibrium constant?
2. Which of the following equilibriums are homogeneous and which are heterogeneous?
[1] The “effective pressure” is called the fugacity, just as activity is the effective concentration.
[2] The reference states for pure solids and liquids are those forms stable at 1 bar (approximately 1 atm), which are
assigned an activity of 1.
15.3 Solving Equilibrium Problems
L E A R N I N G O B JE C T I V E
1. To solve quantitative problems involving chemical equilibriums.
There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the
reactants and the products at equilibrium (or, more often, information that allows us to calculate these
concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are
given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the
concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of
problems.
Calculating an Equilibrium Constant from Equilibrium Concentrations
We saw in the exercise in Example 6 in Section 15.2 "The Equilibrium Constant" that the equilibrium
constant for the decomposition of CaCO3(s) to CaO(s) and CO2(g) is K= [CO2]. At 800°C, the concentration
of CO2 in equilibrium with solid CaCO3 and CaO is 2.5 × 10 − 3
M. Thus K at 800°C is 2.5 × 10 − 3 . (Remember
that equilibrium constants are unitless.)
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