Chapter 14, Problem 14.100QP

Interpretation Introduction

Interpretation:

To show that ${\text{K}}_{\text{c}}\text{=}\frac{{\text{K}}_{\text{p}}}{\text{RT}}$

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Ideal gas equation is an equation that is describing the state of a imaginary ideal gas.

${\text{PV}}_{}\text{=n RT}$

Where,

P is the pressure of the gas

V is the volume

n is the number of moles of gas

R is the universal gas constant $\text{(R=0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

T is the temperature