Ch. 9 Chemical Equilibrium

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dyanmic equilibrium: concentration remains constant, but reactions are occuring at a molecular level
the reaction occurs at the same rate in one direction as the other

Equilibrium Expression

the product of the concentrations of the products divided by the product of the concentrations of the reactants equals the equilibrium constant ( $K$ )
$K_c$ : equilibrium concentration
$K_p$ : equilibrium pressure
$K_sp$ : solubility product
$K_a$ : acid ionization constant
$K_b$ : base ionization constant

In a balanced, simplified equilibrium reaction $\ce{aA + bB <=> pP + nN}$ ,
$K_c=\frac{[P]^p[N]^n}{[A]^a[B]^b}$

Example

What is the equilibrium expression for the combustion of propane ( $\ce{C3H8}$ )?

The balanced equilibrium reaction is $\ce{C3H8 + 5O2 <=> 4H2O + 3CO2}$
So, the equilibrium expression is
$K_c=\frac{[\ce{H2O}]^4[\ce{CO2}]^3}{[\ce{C3H8}][\ce{O2}]^5}$

In general, only aqueous substances (and mixed gases) are included, i.e. solids/liquids are ignored

when equations are added, the equilibrium constants are multiplied
when an equation is reversed, the equillibirum constant is the inverse
when an equation is mutliplied by a constant, the equilibrium constant is raised to the power of that constant

Example

Given the following reactions and equilibrium constants:
$\begin{matrix}\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)} && K_c=5.6\cdot10^9\end{matrix}$
$\begin{matrix}\ce{Ag+(aq) + 2NH3(aq) <=> Ag(NH3)2+(aq)} && K_c=1.6\cdot10^7\end{matrix}$
Calculate the equilibrium constant for
$\ce{AgCl(s) + 2NH3(aq) <=> Ag(NH3)2+(aq) + Cl-(aq)}$

The third equation can be obtained by adding the second equation with the reverse of the first equation.
So, the equilibrium constant desired is $K_c=(1.6\cdot10^7)(\frac{1}{5.6\cdot10^9})=2.9\cdot10^{-3}$

Using the Equilibrium Constant

thermodynamically favorable reaction: reaction occurs when mixed without additional assistance
- $K_c>1.00$
thermodynamically unfavorable reaction: reaction does not occur without additional assistance
- $K_c<1.00$
implies reaction is favored in one direction
reaction quotient ( $Q$ $Q$ ): value of the equilibrium expression when not in equilibrium ( $Q$ $Q$ instead of $K_c$ $K_{c}$ )
- If $Q$ doesn't change over time, equilibrium is reached and $Q=K_c$
- If $Q<K_c$ $Q < K_{c}$ , the reaction will favor the products and move toward the right
  - $Q$ needs to increase, so larger numerator and smaller denominator $\rightarrow$ favor products
- If $Q>K_c$ $Q > K_{c}$ , the reaction will favor the reactants and move toward the left
  - $Q$ needs to decrease, so smaller numerator and larger denominator $\rightarrow$ favor reactants

Example 1

Is the reaction in equilibrium?
$\begin{matrix}\ce{H2(g) + I2(g) <=> 2HI(g)}&&K_c=49\end{matrix}$
$[\ce{H2}]=\pu{0.10 }M$ ; $[\ce{I2}]=\pu{0.1 }M$ ; $[\ce{HI}]=\pu{0.70 }M$

Calculate the reaction quotient:
$Q=\frac{0.70^2}{0.1\cdot0.1}=49$
Since $Q=K_c$ , the reaction is in equilibrium.

Example 2

Which direction will the following reaction go in?
$\begin{matrix}\ce{HF(aq) + H2O(l) <=> F-(aq) + H3O+(aq)}&&K_c=6.8\cdot10^{-4}\end{matrix}$
$[\ce{HF}]=\pu{0.20 }M$ ; $[\ce{F-}]=\pu{2.0*10^{-4} } M$ ; $[\ce{H3O+}]=\pu{2.0*10^-4 }M$

Calculate the reaction quotient, noting that water is ignored
$Q=\frac{(2.0\cdot10^{-4})(2.0\cdot10^{-4})}{0.20}=2.0\cdot10^{-7}$
Since $Q<K_c$ , the reaction favors the products and move right

RICE Tables

method of organizing equilibrium concentrations and make calculations
Reaction: the chemical reaction
Initial concentration: before reactions occur
Change: the change in concentration, usually written in terms of $x$ and follows mole ratio
Equillibrium: expression for the concentration of the substance at equilibrium; usually written in terms of $x$

Example 1

If $\pu{0.250 } M$ each of $\ce{NO2}$ , $\ce{SO2}$ , $\ce{NO}$ , and $\ce{SO3}$ are mixed until equilibrium is reached, $\pu{0.261 } M$ of $\ce{NO2}$
was measured in the container. What is the equilibrium constant for this reaction?

We start the RICE table with the balanced reaction

Reaction	$\ce{NO2(g)}$	$+$	$\ce{SO2(g)}$	$\ce{<=>}$	$\ce{NO(g)}$	$+$	$\ce{SO3(g)}$
Initial
Change
Equilibrium

Next we fill in the known information, including initial concentration, the change, and the equilibrium expressions

Reaction	$\ce{NO2(g)}$	$\ce{SO2(g)}$	$\ce{NO(g)}$	$\ce{SO3(g)}$
Initial	$\pu{0.250 }M$	$\pu{0.250 }M$	$\pu{0.250 }M$	$\pu{0.250 }M$
Change	$+x$	$+x$	$-x$	$-x$
Equilibrium	$0.250+x$	$0.250+x$	$0.250-x$	$0.250-x$

Since the equilibrium concentration of $\ce{NO2}$ is $\pu{0.261 }M$ ,
$0.250+x=0.261\rightarrow x=0.011$
So, the equilibrium constant is
$K_c=\frac{(0.250-x)(0.250-x)}{(0.250+x)(0.250+x)}=\frac{0.239^2}{0.261^2}=0.839$

Example 2

A container is filled with $\pu{0.200 }M$ $\ce{BrCl}$ . Given the following equation, what is the equilibrium concentration of $\ce{Br2}$ and $\ce{Cl2}$ ?
$\begin{matrix}\ce{Br2 + Cl2 <=> 2BrCl}&&K_c=6.90\end{matrix}$

Create the RICE table

Reaction	$\ce{Br2}$	$\ce{Cl2}$	$\ce{2BrCl}$
Initial	$\pu{0 }M$	$\pu{0 }M$	$\pu{0.200 }M$
Change	$+x$	$+x$	$-2x$
Equilibrium	$x$	$x$	$0.200-2x$

Now, we can make the equilibrium expression
$K_c=6.90=\frac{(0.200-2x)^2}{x^2}\rightarrow 2.627=\frac{0.200-2x}{x}\rightarrow x=0.0432$
So, the final concentrations are $[\ce{Br2}]=\pu{0.0432 }M$ , $[\ce{Cl2}]=\pu{0.0432 }M$ , and $[\ce{BrCl}]=\pu{0.114 }M$

If $K_c$ is very small, $x$ may be ignored where it is added or subtracted due to approximations

Example

If $\pu{2.00 moles}$ of $\ce{SO3}$ is placed in a $\pu{1.00 L}$ flask, what are the final concentrations of the substances in the flask?
$\begin{matrix}\ce{2SO3(g) <=> 2SO2(g) + O2(g)}&&K_c=2.4\cdot10^{-25}\end{matrix}$

Create the RICE table, noting that $[\ce{SO3}]=\frac{\pu{2.00 mol}}{\pu{1.00L}}=\pu{2.00 }M$

Reaction	$\ce{2SO3}$	$\ce{2SO2}$	$\ce{O2}$
Initial	$\pu{2.00 }M$	$\pu{0 }M$	$\pu{0 }M$
Change	$-2x$	$+2x$	$+x$
Equilibrium	$2.00-2x$	$2x$	$x$

Plugging into the equilibrium expression,
$K_c=2.4\cdot10^{-25}=\frac{x(2x)^2}{(2.00-2x)^2}$
Since $K_c$ is so small, it can be assumed that $x$ is also very small, so $2.00-2x\approx 2.00$ , making the equation
$K_c=2.4\cdot10^{-25}=\frac{x(2x)^2}{(2.00)^2}\rightarrow x^3=2.4\cdot10^{-25}\rightarrow x=6.2\cdot10^{-9}$
So, the final concentrations are $[\ce{SO3}]=\pu{2.00 }M$ , $[\ce{SO2}]=1.2\cdot10^{-8}$ , and $[\ce{O2}]=6.2\cdot10^{-9}$

Since $PV=nRT$ can be rearranged to be $\frac{n}{V}=\frac{P}{RT}$ , the molar concentration is proportional to the pressure under a constant temperature.
So, the same equation and process may be used with partial pressures and the equilibrium constant $K_p$
In general, $K_p=K_c(RT)^{\Delta n_g}$
where $K_p$ is equilibrium pressure, $K_c$ is equilibrium concentration, $R$ is the ideal gas constant, $T$ is the Kelvin temperature, and $\Delta n_g$ is the change in the number of moles of gas in the balanced reaction

Solubility Product

In determining if a solid is soluble, we use $K_{sp}$
Usually the equation looks like $\ce{aA(s) <=> bB(aq) + cC(aq)}$ , so $K_{sp}=[B]^b[C]^c$

molar solubility ( $\pu{mol/L}$ $m o l / L$ ) of a compound is how much of the substance dissolves
- equals $x$

Example

If the molar solubility of $\ce{MgF2}$ is $\pu{1.18*10^-3 mol/L}$ , what is $K_{sp}$ ?

Set up the RICE table

Reaction	$\ce{MgF2}$	$\ce{Mg^2+}$	$\ce{2F-}$
Initial	Solid	$\pu{0 }M$	$\pu{0 }M$
Change	-	$+x$	$+2x$
Equilibrium	-	$x$	$2x$

We know the molar solubility is $x=1.18\cdot10^{-3}$ , so the expression for $K_{sp}$ is
$K_{sp}=[\ce{Mg^2+}][\ce{F-}]^2=4x^3=4(1.18\cdot10^{-3})^3=6.57\cdot10^{-9}$

Weak-Acid/Weak-Base Equilibria

Weak acids and bases ionize partly when dissolved in water

$\ce{CH3COOH}$ is an acid, and dissolution can be expressed as $\ce{CH3COOH(aq) + H2O(l) <=> CH3COO-(aq) + H3O+(aq)}$ and the equilibrium constant is $K_c=K_a=\frac{[\ce{CH3COO-}][\ce{H3O+}]}{[\ce{CH3COOH}]}$
$\ce{NH3}$ is a base, and dissolution can be expressed as $\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}$ and the equilibrium constant is $K_c=K_b=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$
$K_a$ is the acid ionization constant
$K_b$ is the base ionization constant

Le Châtlier's Principle

Le Châtelier's Principle: when a system in dynamic equilibrium is disturbed, it will respond to reestablish a new equilibrium state, if possible
- changing concentration affects $Q$ , so favors one side
- changing pressure for gases increases molar concentration due to changed volume
- changing temperature is the only thing that can change $K$
  - depends on whether reaction is exothermic or endothermic, and affects reaction