# 13.2 Equilibrium constants  (Page 5/14)

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The equation relating K c and K P is derived as follows. For the gas-phase reaction $m\text{A}+n\text{B}⇌x\text{C}+y\text{D:}$

${K}_{P}=\phantom{\rule{0.2em}{0ex}}\frac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}$
$=\phantom{\rule{0.2em}{0ex}}\frac{{\left(\left[\text{C]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT\right)}^{x}\left(\left[\text{D]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT{\right)}^{y}}{\left(\left[\text{A]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT{\right)}^{m}{\left(\left[\text{B]}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}RT\right)}^{n}}$
$=\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{C}{\right]}^{x}{\left[\text{D}\right]}^{y}}{\left[\text{A}{\right]}^{m}{\left[\text{B}\right]}^{n}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{{\left(RT\right)}^{x\text{+}y}}{{\left(RT\right)}^{m\text{+}n}}$
$={K}_{c}{\left(RT\right)}^{\left(x\text{+}y\right)-\left(m\text{+}n\right)}$
$={K}_{c}{\left(RT\right)}^{\text{Δ}n}$

The relationship between K c and K P is

${K}_{P}={K}_{c}{\left(RT\right)}^{\text{Δ}n}$

In this equation, Δ n is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction $m\text{A}+n\text{B}⇌x\text{C}+y\text{D,}$ we have

$\text{Δ}n=\left(x\text{+}y\right)-\left(m\text{+}n\right)$

## Calculation of K P

Write the equations for the conversion of K c to K P for each of the following reactions:

(a) ${\text{C}}_{2}{\text{H}}_{6}\left(g\right)⇌{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)$

(b) $\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)$

(c) ${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)$

(d) K c is equal to 0.28 for the following reaction at 900 °C:

${\text{CS}}_{2}\left(g\right)+4{\text{H}}_{2}\left(g\right)⇌{\text{CH}}_{4}\left(g\right)+2{\text{H}}_{2}\text{S}\left(g\right)$

What is K P at this temperature?

## Solution

(a) Δ n = (2) − (1) = 1
K P = K c ( RT ) Δ n = K c ( RT ) 1 = K c ( RT )

(b) Δ n = (2) − (2) = 0
K P = K c ( RT ) Δ n = K c ( RT ) 0 = K c

(c) Δ n = (2) − (1 + 3) = −2
K P = K c ( RT ) Δ n = K c ( RT ) −2 = $\frac{{K}_{c}}{{\left(RT\right)}^{2}}$

(d) K P = K c (RT) Δ n = (0.28)[(0.0821)(1173)] −2 = 3.0 $×$ 10 −5

Write the equations for the conversion of K c to K P for each of the following reactions, which occur in the gas phase:

(a) $2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{SO}}_{3}\left(g\right)$

(b) ${\text{N}}_{2}{\text{O}}_{4}\left(g\right)⇌2{\text{NO}}_{2}\left(g\right)$

(c) ${\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)⇌3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)$

(d) At 227 °C, the following reaction has K c = 0.0952:

${\text{CH}}_{3}\text{OH}\left(g\right)⇌\text{CO}\left(g\right)+2{\text{H}}_{2}\left(g\right)$

What would be the value of K P at this temperature?

(a) K P = K c ( RT ) −1 ; (b) K P = K c ( RT ); (c) K P = K c ( RT ); (d) 160 or 1.6 $×$ 10 2

## Heterogeneous equilibria

A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).

Some heterogeneous equilibria involve chemical changes; for example:

${\text{PbCl}}_{2}\left(s\right)⇌{\text{Pb}}^{2+}\left(aq\right)+2{\text{Cl}}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{\text{−}}\right]}^{2}$
$\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)⇌{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{1}{{\text{[CO}}_{2}\right]}$
$\text{C}\left(s\right)+2\text{S}\left(g\right)⇌{\text{CS}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\text{[CS}}_{2}\right]}{{\left[\text{S}\right]}^{2}}$

Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:

${\text{Br}}_{2}\left(l\right)⇌{\text{Br}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\left[{\text{Br}}_{2}\right]$

We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:

$\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)⇌{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{5em}{0ex}}{K}_{P}=\phantom{\rule{0.2em}{0ex}}\frac{1}{{P}_{{\text{CO}}_{2}}}$
$\text{C}\left(s\right)+2\text{S}\left(g\right)⇌{\text{CS}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{P}=\phantom{\rule{0.2em}{0ex}}\frac{{P}_{{\text{CS}}_{2}}}{{\left({P}_{\text{S}}\right)}^{2}}$

## Key concepts and summary

For any reaction that is at equilibrium, the reaction quotient Q is equal to the equilibrium constant K for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant. Their particular values may vary depending on conditions, but the value of the reaction quotient will always equal K ( K c when using concentrations or K P when using partial pressures).

A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases. We can decide whether a reaction is at equilibrium by comparing the reaction quotient with the equilibrium constant for the reaction.

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