11.1-11.2 - Kc And Kp Calculations

Why An Equilibrium Constant Is Useful

Imagine sealing nitrogen dioxide and dinitrogen tetroxide in a flask:

The mixture is dynamic: molecules keep reacting in both directions. At equilibrium the forward and reverse rates are equal, but the amounts of each gas do not have to be equal. The useful question is not "has the reaction stopped?" but "what ratio of products to reactants is fixed at this temperature?"

For this equation, the pressure ratio

has a constant value at a fixed temperature once equilibrium has been reached. A different starting mixture can reach the same equilibrium constant because the constant describes the final equilibrium ratio for the equation, not the route taken to get there.

There are two common forms in this specification:

• $K_{\mathrm c}$ uses equilibrium concentrations, usually in $\mathrm{mol\,dm^{-3}}$.
• $K_{\mathrm p}$ uses equilibrium partial pressures of gases, in atm.

For the same defined equation and unit convention, a larger equilibrium constant means a larger product-to-reactant concentration or partial-pressure ratio, with the stoichiometric powers included. Actual product fractions also depend on the total concentration or pressure and the starting mixture; a dimensional constant alone is not a universal percentage yield. Do not treat this as a rate statement: $K_{\mathrm c}$ and $K_{\mathrm p}$ describe position at equilibrium, not how quickly equilibrium is reached.

Writing Kc And Kp Expressions

Start with the balanced equation. Products go on the top, reactants go on the bottom, and the powers come from the coefficients in that exact equation.

For a general homogeneous equilibrium:

Homogeneous examples

All species are in the same phase:

For the gas equilibrium

A gas equilibrium can also be treated with $K_{\mathrm c}$ if the data is given as concentrations:

Heterogeneous examples

A heterogeneous equilibrium contains more than one phase. In A level equilibrium expressions, pure solids and pure liquids are omitted because their effective concentration is treated as constant.

The solid carbon is not included:

If the same idea is expressed as a concentration constant for a heterogeneous reaction, pure solids are still omitted:

For thermal decomposition:

both solids are omitted:

$K_{\mathrm p} = p(\ce{CO2})$

Common-error contrast

Wrong:

This treats solids as if they had partial pressures. The correct expression is only $p(\ce{CO2})$.

Calculating Kc From Experimental Data

$K_{\mathrm c}$ calculations need equilibrium concentrations. If the data gives amounts in moles and a volume, convert each equilibrium amount using:

$c = n/V$, with $n$ in mol and $V$ in $\mathrm{dm^{3}}$.

Worked example: an equilibrium amount table

Hydrogen and iodine react in a sealed $2.00 \mathrm{dm^{3}}$ container:

Initially there are $0.300\,\mathrm{mol}$ of $\ce{H2}$ and $0.300\,\mathrm{mol}$ of $\ce{I2}$, with no $\ce{HI}$. At equilibrium, $0.400\,\mathrm{mol}$ of $\ce{HI}$ is present. Calculate $K_{\mathrm c}$.

Step 1: use the mole ratio.

For every $2\,\mathrm{mol}$ of $\ce{HI}$ formed, $1\,\mathrm{mol}$ of $\ce{H2}$ and $1\,\mathrm{mol}$ of $\ce{I2}$ are used.

$\ce{HI}$ formed $= 0.400\,\mathrm{mol}$, so $\ce{H2}$ used $= 0.200\,\mathrm{mol}$ and $\ce{I2}$ used $= 0.200\,\mathrm{mol}$.

Step 2: find equilibrium amounts.

Step 3: convert to equilibrium concentrations.

Step 4: substitute into the expression.

$K_{\mathrm c} = 16.0$

Step 5: check units.

The units cancel, so this $K_{\mathrm c}$ has no units.

Chemical interpretation: for this fixed temperature and the equation as written, the equilibrium concentration ratio favours $\ce{HI}$ compared with the remaining $\ce{H2}$ and $\ce{I2}$.