Chapter 19, Problem 19.1P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ${\text{[SbCl}}_{\text{3}}\text{]}$ and ${\text{[Cl}}_{\text{2}}\text{]}$ for the given reaction, ${\text{SbCl}}_{\text{5(g)}}\stackrel{}{\rightleftharpoons }{\text{SbCl}}_{\text{3(g)}}{\text{ + Cl}}_{\text{2(g)}}$ has to be calculated.

Concept Introduction:

Equilibrium constant: At equilibrium the ratio of products to reactants (each raised to the power corresponding to its stoichiometric coefficient) has a constant value, K

For a general reaction,

  $\text{aA+bB}\rightleftarrows \text{cC+dD}$

  $\text{Equilibrium}\text{constant}\text{K}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

The solids and pure liquids are not involved in equilibrium constant expression.

Explanation of Solution

Given:

Antimony pentachloride decomposes according to the below equation.

  ${\text{SbCl}}_{\text{5(g)}}\stackrel{}{\rightleftharpoons }{\text{SbCl}}_{\text{3(g)}}{\text{ + Cl}}_{\text{2(g)}}$

The initial concentrations of all the reactant and products are,

  $\begin{array}{l}{{\text{[SbCl}}_{\text{5}}}_{\text{]}}\text{ = 0}\text{.165 M}\\ {{\text{[SbCl}}_{\text{3}}}_{\text{]}}\text{ = 0}\text{.0955 M}\\ {\text{ [Cl2]}}_{\text{0}}\text{ = 0}\text{.210 M}\end{array}$

The concentration of ${\text{[SbCl}}_{\text{5}}\text{]}$ at equilibrium is $\text{0}\text{.135 M}$.

Calculation of equilibrium concentrations:

Consider ‘x’ moles of ${\text{[SbCl}}_{\text{5}}\text{]}$ decomposes and gets converted to ${\text{[SbCl}}_{\text{3}}\text{]}$ and ${\text{[Cl}}_{\text{2}}\text{]}$ at equilibrium.

	${\text{SbCl}}_{\text{5(g)}}\stackrel{}{\rightleftharpoons }{\text{ SbCl}}_{\text{3(g)}}{\text{ + Cl}}_{\text{2(g)}}$
Initial $\text{(M)}$	$\text{0}\text{.165 M}$	$\text{0}\text{.0955 M}$	$\text{0}\text{.210 M}$
Change $\text{(M)}$	-x	+x	+x
Equilibrium $\text{(M)}$	$\text{0}\text{.165 M-x}$	$\text{0}\text{.0955 M+x}$	$\text{0}\text{.210 M+x}$

Solve for x using equilibrium concentration of ${\text{[SbCl}}_{\text{5}}\text{]=0}\text{.135 M}$

  $\begin{array}{l}{{\text{[SbCl}}_{\text{5}}}_{\text{]}}\text{ = 0}\text{.135 M}\\ \text{0}\text{.165 M - x = 0}\text{.135 M}\\ \text{x =0}\text{.165 M - 0}\text{.135 M}\\ \text{ = 0}\text{.03 M}\end{array}$

Now calculate the equilibrium concentrations of ${\text{[SbCl}}_{\text{3}}\text{]}$ and ${\text{[Cl}}_{\text{2}}\text{]}$ as below.

  $\begin{array}{l}\text{ x = 0}\text{.03 M}\\ {{\text{[SbCl}}_3}_{\text{]}}=\text{0}\text{.0955 M+x}\\ \text{ =0}\text{.0955 M+ 0}\text{.03 M}\\ \text{ = 0}\text{.1255 M}\\ {\text{ [Cl}}_2{\text{]}}_{\text{eq}}=\text{0}\text{.210 M+x}\\ \text{ =0}\text{.210 M+ 0}\text{.03 M}\\ \text{ = 0}\text{.24 M}\end{array}$

Therefore, the equilibrium concentrations of ${\text{[SbCl}}_{\text{3}}\text{]}$ and ${\text{[Cl}}_{\text{2}}\text{]}$ are $\text{0}\text{.1255 M}$ and $\text{0}\text{.24 M}$ respectively.