Equilibrium (II)

$K_c$ expressions

It is a fixed value for a specific mixture & temperature

There are two types of mixture:

Homogenous – all components in same state

\begin{align}aA+bB\ \rightleftharpoons\ cC+dD\end{align} and at a constant temperature Kc is shown by: \begin{align}\frac{\left[C\right]^c\ast\left[D\right]^d}{\left[A\right]^a\ast\left[B\right]^b}=\ K_c\end{align} For example: \begin{align}K_c=\left[\frac{\left[NH_3\right]^2}{\left[N_2\right]^1\ast\left[H_2\right]^3}\right]\end{align}

Heterogenous – components in one or more states

In this case, solid or pure liquid reactants or products are not included in the expression. If water is present in large amounts discard, else include

The difficult types of question

$A+3B\rightleftharpoons C\ +\ D$

Equation for reaction Reactants Products
Starting amount 5 4 0 0
Equilibrium amount 0.82
Ratio 1 3 1 1
Amount that reacts 0.82 3 * 0.82 - -
Amount that forms - - 0.82 0.82
Eq. amounts 5-0.82 4-2.46 0.82 0.82
Concentration/partial pressure 4.18/0.50 1.54/0.50 0.82/0.50 0.82/0.50

$H_2+I_2\rightleftharpoons 2HI$

$H_2$ $I_2$ $2HI$
Initial 1.9 1.9 0
Equilibrium 3
Amount lost 1.5 1.5 -
At equilibrium 1.9-1.5 = 0.4 1.9-1.5 = 0.4 3
Concentration 0.4/0.25 0.4/0.25 3/0.25
Then substitute back into expression

The Equilibrium constant for gases: $K_p$

What is partial pressure?
It’s difficult to measure the concentration of gases. Instead, the quantity of each gas is described by the pressure it exerts – the partial pressure
Defined as the pressure of one of the species
If number of moles is not given, use those shown in the balanced equation

\begin{align}partial\ pressure\ =\ mole\ fraction\ \ast\ total\ pressure\end{align} \begin{align}p_A=x_A*p_{total}\end{align}

A mole fraction is given by: \begin{align}mole\ fraction\ of\ A=\frac{no.\ moles\ of\ species\ A}{total\ no.\ moles\ of\ all\ species}\end{align}

For general equilibrium: \begin{align}K_p=\frac{p\left(C\right)^cp\left(D\right)^d}{p\left(A\right)^ap\left(B\right)^b}\end{align}

Like $K_C$, we must work out the units. The units are either $Pa$ or a derivative. \begin{align}K_p=\frac{{kPa}^c{kPa}^d}{{kPa}^a{kPa}^b}\end{align}

Let’s try: 2.0 moles $H_2$ & 1.0 moles $CO$ were mixed together. At equilibrium, the mixture contained 0.175 moles of $CH_3OH$. Work out $K_p$ using the formula $2H_2+CO \rightleftharpoons CH_3OH$:

$H_2$ $CO$ $CH_3OH$
Moles at start 2 1 0
Moles reacted 0.35 0.175 -
Moles at equilibrium 1.65 0.825 0.175
Mole fraction 0.623 0.311 0.066
Partial pressure 10902 5442.5 1155

Factors affecting equilibria

The only thing to affect $K_c$ & $K_p$ is temperature. Le Chatelier’s principle.

• The effect depends on whether the reaction is exothermic or endothermic
• If exothermic, increasing temperature shifts to the left, $K_c$ decreases
• If endothermic, increasing temperature shifts to the right, $K_c$ increases
• If concentration of one increases, the concentration in the whole system will change to keep $K_c$ constant, because one side of the reaction is favoured
• If pressure changes, the concentration changes automatically.
• Catalysts doesn’t affect constants, just the rate of reaction. They increase the rate of forward & backward