Question

Explain all the terms in this equation.
\[
  {K_P} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}{\left( {RT} \right)^{\left( {{n_p} - {n_R}} \right)}} \\
   = {K_c}{\left( {RT} \right)^{\vartriangle ng}} \\
 \]

Answer 1

Hint: ${K_c}$ and ${K_p}$ are the equilibrium constants of gaseous mixtures. However, the difference between the constants is that ${K_c}$ is defined by molar concentrations, where ${K_p}$ is defined by the partial pressures of the gases inside a closed system. The equilibrium constants do not include the concentrations of single components such as liquids and solids and they may have units of the reaction.

Complete step by step answer:
When forward and reverse reactions occur at the same rate, the system reaches dynamic equilibrium. Chemical equilibrium occurs when dynamic equilibrium realizes for all steps of the reaction. At a given temperature, the equilibrium composition is related to the equilibrium constant ${K_c}$. For the general reaction \[aA + bB \to cC + dD\]. ${K_c}$ is the ratio of the product concentrations with each concentration raised to the power of its coefficient in the balanced chemical reaction?
 $aA + bB \rightleftharpoons cC + dD$
 ${K_p} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$
The relationship between the two equilibrium constants are:
 $
  {K_p} = {K_c}{\left( {RT} \right)^{\vartriangle n}} \\
  or\,{K_c} = \dfrac{{{K_p}}}{{{{\left( {RT} \right)}^{\vartriangle n}}}} \\
 $
Regardless of the initial concentrations the final equilibrium concentrations must satisfy the equation specified by ${K_c}$. Usually ${K_c}$ is written without units.
If an equilibrium involves reactants and products in a single phase that is called a homogeneous equilibrium opposite to a heterogeneous equilibrium which involves reactants and products in more than one phase.

Note: When equilibrium constant for a reaction needs to be calculated from equilibrium constants of individual steps one should take an approach analogous of the use of Hess’s law. As the enthalpy change for the overall equation equals to the sum of enthalpy changes for the individual steps.