Chapter 12, Problem 97QRT

(a)

Interpretation Introduction

Interpretation:

For the formation of carbonylbromide from Carbon monoxide and Bromine, the value of ${\text{K}}_{\text{P}}$ has to be calculated.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{p}}\right)$:

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration.  Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$ is defined as ratio of partial pressure of products to partial pressure of reactants.  Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

    $\text{aA}\rightleftharpoons \text{bB}$

    $\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\text{P}}_{\text{A}}{}^{\text{a}}{\text{=k}}_{\text{r}}{\text{P}}_{\text{B}}{}^{\text{a}}\end{array}$

On rearranging,

    $\frac{{\text{P}}_{\text{B}}{}^{\text{b}}}{{\text{P}}_{\text{A}}{}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{p}}$

Where,

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

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

    ${\text{K}}_{\text{p}}$ is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

  $\begin{array}{l}{\text{P}}_{\text{A}}{\text{V = n}}_{\text{A}}\text{RT}\\ {\text{P}}_{\text{A}}\text{ = }\frac{{\text{n}}_{\text{A}}}{\text{V}}\text{RT}\text{=}\text{[A]RT}\end{array}$

In same way ${\text{P}}_{\text{B}}\text{ =}\text{[B]RT}$.  Thus, equilibrium constant ${\text{K}}_{\text{p}}$ is given as

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{=}\frac{{\text{[B]}}^{\text{b}}}{{\text{[A]}}^{\text{a}}}\text{×}{\text{(RT)}}^{\text{b-a}}\\ {\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\text{Δn}}\end{array}$

Where,

  R is gas constant,

  T is absolute temperature,

  $\text{Δn}$ is difference in moles of products to reactants in gas phase.

Explanation of Solution

Given information,

  Formation of carbonylbromide from Carbon monoxide and Bromine,

  ${\text{CO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{COBr}}_{\text{2}\left(\text{g}\right)}$

Equilibrium temperature is $\text{346}\text{K}$, the partial pressures of ${\text{COBr}}_{\text{2}}\text{,}\text{CO}\text{,}\text{and}{\text{Br}}_{\text{2}}$ are 0.12, 1.00, and $\text{0}\text{.65}\text{atm}$ respectively.

Calculate the ${\text{K}}_{\text{p}}$ value for the given reaction

  $\begin{array}{l}{\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}_{{\text{COBr}}_{\text{2}}}}{{\text{P}}_{\text{CO}}{\text{P}}_{{\text{Br}}_{\text{2}}}}\text{=}\frac{\text{0}\text{.12}}{\left(\text{1}\text{.0}\right)\left(\text{0}\text{.65}\right)}\\ \\ \text{=}\text{0}\text{.1846}\text{=}\text{0}\text{.18}\end{array}$

Therefore, the ${\text{K}}_{\text{p}}$ value for the given reaction is 0.18.

(b)

Interpretation Introduction

Interpretation:

In the formation of carbonylbromide from Carbon monoxide and Bromine, the partial pressure of Bromine is decreased into 0.50, now the partial pressure of all gases has to be calculated after equilibrium is re-established.

Concept Introduction:

Le Chatelier's principle:

It states that if a system in equilibrium gets disturbed due to modification of concentration, temperature, volume, and pressure, then it reset to counteract the effect of disturbance.

Explanation of Solution

Given information,

  Formation of carbonylbromide from Carbon monoxide and Bromine,

  ${\text{CO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{COBr}}_{\text{2}\left(\text{g}\right)}$

Equilibrium temperature is $\text{346}\text{K}$, the partial pressures of ${\text{COBr}}_{\text{2}}\text{,}\text{CO}\text{,}\text{and}{\text{Br}}_{\text{2}}$ are 0.12, 1.00, and $\text{0}\text{.50}\text{atm}$ respectively.

Construct ICE table for the above reaction,

  $\begin{array}{l}\underset{\_}{{\text{CO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{COBr}}_{\text{2}\left(\text{g}\right)}}\\ \text{Initial}\text{pressure}\text{(atm):}\text{1}\text{.00}\text{0}\text{.50}\text{0}\text{.12}\\ \text{Change}\text{in}\text{press}\text{.}\text{(atm):}\text{+x}\text{+x}\text{-x}\\ \text{Equilibrium}\text{pressure}\text{(atm):}\text{1}\text{.00}\text{+}\text{x}\text{0}\text{.50}\text{+}\text{x}\text{0}\text{.12}\text{-}\text{x}\end{array}$

At equilibrium,

  ${\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}_{{\text{COBr}}_{\text{2}}}}{{\text{P}}_{\text{CO}}{\text{P}}_{{\text{Br}}_{\text{2}}}}\text{=}\frac{\text{0}\text{.12}\text{-}\text{x}}{\left(\text{1}\text{.0}\text{+}\text{x}\right)\left(\text{0}\text{.50}\text{+}\text{x}\right)}\text{=}\text{0}\text{.1846}$

Construct quadratic equation,

$\begin{array}{l}\left(\text{0}\text{.1846}\right)\left(\text{1}\text{.00}\right)\left(\text{0}\text{.50}\right)\text{ + }\left(\text{0}\text{.1846}\right)\left(\text{1}\text{.00}\right)\text{x + }\left(\text{0}\text{.1846}\right)\left(\text{0}\text{.50}\right)\text{x + }\left(\text{0}\text{.1846}\right){\text{x}}^{\text{2}}\text{ = 0}\text{.12-}\text{x}\\ \\ \left(\text{0}\text{.09231}\right)\text{+}\left(\text{0}\text{.1846}\text{+}\text{0}\text{.09231}\right)\text{x}\text{+}\left(\text{0}\text{.1846}\right){\text{x}}^{\text{2}}\text{=}\text{0}\text{.12}\text{-}\text{x}\\ \\ \text{0}{\text{.1846x}}^{\text{2}}\text{+}\text{1}\text{.2769}\text{x}\text{-}\text{0}\text{.02769}\text{=}\text{0}\end{array}$

Find the value of x,

  $\begin{array}{l}\text{x}\text{=}\frac{\text{-}\text{b}\text{±}\sqrt{{\text{b}}^{\text{2}}\text{-}\text{4}\text{ac}}}{\text{2}\text{a}}\\ \\ \text{=}\frac{\text{-}\text{1}\text{.2769}\text{±}\sqrt{{\left(\text{1}\text{.2769}\right)}^{\text{2}}\text{-}\text{4}\left(\text{0}\text{.1846}\right)\left(\text{-0}\text{.0276}\right)}}{\text{2}\left(\text{0}\text{.1846}\right)}\\ \\ \text{=}\frac{\text{-}\text{1}\text{.2769}\text{±}\sqrt{\text{1}\text{.6305}\text{+}\text{0}\text{.0204}}}{\text{0}\text{.3692}}\text{=}\frac{\text{-}\text{1}\text{.2769}\text{+}\text{1}\text{.2849}}{\text{0}\text{.3692}}\\ \\ \text{=}\text{0}\text{.0216}\end{array}$

Calculate the partial pressure of all gases by using the value of x,

  $\begin{array}{l}{\text{P}}_{\text{CO}}\text{=}\text{1}\text{.00}\text{+}\text{0}\text{.02}\text{=}\text{1}\text{.02}\text{atm}\\ \\ {\text{P}}_{{\text{H}}_{\text{2}}}\text{=}\text{0}\text{.50}\text{+}\text{0}\text{.02}\text{=}\text{0}\text{.52}\text{atm}\\ \\ {\text{P}}_{{\text{COBr}}_{\text{2}}}\text{=}\text{0}\text{.12}\text{-}\text{0}\text{.02}\text{=}\text{0}\text{.10}\text{atm}\end{array}$