Chapter 8, Problem 118QRT

(a)

Interpretation Introduction

Interpretation:

$PC{l}_5$ decomposition whether takes place has to be determined using calculation.

Concept Introduction:

Molar mass can be determined using the given equation,

  $\text{n}\text{= }\frac{\text{Mass}}{\text{Molar mass}}$

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\text{PV = nRT}$

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Explanation of Solution

Number of moles of $PC{l}_5$ before vaporization is calculated as given below,

  $\begin{array}{c}n\text{ = }\frac{Mass}{Molar\text{ mass}}\\ \\ =\frac{\text{2}\text{.69 g}}{\text{208}\text{.22 g/mol}}\\ \\ =\text{ 0}\text{.0129 mol}\end{array}$

Number of moles present after vaporization is calculated using Ideal Gas Lew as given below,

  $\begin{array}{c}T\text{}\text{}\text{ = 250}°\text{C = }\left(250+273\right)K=\text{ 523 K}\\ \\ PV\text{}\text{ = nRT}\\ \\ \left(1.0\text{ atm}\right)\left(1.0\text{ L}\right)=\text{ n}\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(\text{523 K}\right)\\ \\ n\text{ = }\frac{\left(1.0\text{ atm}\right)\left(1.0\text{ L}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(\text{523 K}\right)}\\ \\ \text{= 0}\text{.0233 mol}\end{array}$

Number of moles before and after vaporization are not equal. Hence, decomposition of $PC{l}_5$ takes place.

(b)

Interpretation Introduction

Interpretation:

Partial pressure of each of the gaseous compound has to be calculated.

Concept Introduction:

Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

  ${\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}$

Answer to Problem 118QRT

Partial pressure of each of the gaseous compound is given below,

  $\begin{array}{c}{p}_{PC{l}_5}\text{= 0}\text{.11 atm}\\ \\ p_{PC{l}_3}\text{= 0}\text{.446 atm}\\ \\ p_{C{l}_2}\text{ = 0}\text{.446 atm}\end{array}$

Explanation of Solution

Total number of moles is $\text{0}\text{.0233 mol}\text{.}$

Initial number of moles of $PC{l}_5$ is $\text{0}\text{.0129 mol}\text{.}$

Change in number of moles of reactants and products can be determined as given below,

  $\begin{array}{l}{\text{ PCl}}_5\underset{}{\to }{\text{ PCl}}_3{\text{ + Cl}}_2\\ \begin{array}{cccc}\text{Initial moles}& 0.0129& \text{0 }& \text{ 0 }\\ Change\text{ in n}& -x& +x& +x\\ \text{Final moles}& \left(0.0129-x\right)& x& x\end{array}\end{array}$

According to Dalton’s law, the total pressure exerted by each gas in a mixture is equal to the sum of the individual partial pressure of the gases.

  $\begin{array}{c}{\text{P}}_{\text{total}}{\text{ = p}}_{PC{l}_5}{\text{ + p}}_{PC{l}_3}{\text{ + p}}_{{\text{Cl}}_{\text{2}}}\\ \\ \left(\text{0}\text{.0233 mol}\right)=\left(\text{0}\text{.0129 mol}-x\right)+x+x\\ \\ x\text{ = 0}{\text{.0104 mol = p}}_{PC{l}_3}{\text{ = p}}_{{\text{Cl}}_{\text{2}}}\\ \\ {\text{p}}_{PC{l}_5}\text{ = }\left(\text{0}\text{.0129 mol}-x\right)\\ \\ =\left(\text{0}\text{.0129 mol}-\text{0}\text{.0104}\right)\\ \\ =\text{ 2}\text{.5}\times {\text{10}}^{-3}\text{ mol}\end{array}$

Mole fraction of each of the gaseous compound is calculated as follows,

  $\begin{array}{c}{X}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}\\ \\ X_{PC{l}_3}\text{=}\frac{\text{0}\text{.0104 mol}}{\text{0}\text{.0233 mol}}\\ \\ =\text{ 0}\text{.446}\\ \\ X_{PC{l}_3}\text{= }{X}_{C{l}_2}\text{= 0}\text{.446}\\ \\ X_{PC{l}_5}\text{=}\frac{\text{2}\text{.5}\times {\text{10}}^{-3}\text{ mol}}{\text{0}\text{.0233 mol}}\\ \\ =\text{ 0}\text{.11}\end{array}$

Partial pressure of each of the gaseous compound is determined as follows,

  $\begin{array}{c}{\text{p}}_{\text{a}}\text{=}{X}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \\ p_{PC{l}_5}\text{= }{X}_{PC{l}_5}\text{×}{\text{P}}_{\text{total}}\\ \\ =\text{ 0}\text{.11}\times \text{1}\text{.00 atm = 0}\text{.11 atm}\\ \\ p_{PC{l}_3}\text{= }{X}_{PC{l}_3}\text{×}{\text{P}}_{\text{total}}\\ \\ =\text{ 0}\text{.446}\times \text{1}\text{.00 atm = 0}\text{.446 atm}\\ \\ p_{C{l}_2}\text{= }{X}_{C{l}_2}\text{×}{\text{P}}_{\text{total}}\\ \\ =\text{ 0}\text{.446}\times \text{1}\text{.00 atm = 0}\text{.446 atm}\end{array}$