Chapter 4, Problem 4B.9P

(a)

Interpretation Introduction

Interpretation: The normal boiling point of nitric acid has to be calculated.

Concept introduction: The normal boiling point is calculated by using the Clausius-Clapeyron’s equation.  The Clausius-Clapeyron’s equation shows the correlation between temperatures and pressures for the phase transition process.

Answer to Problem 4B.9P

The normal boiling point of nitric acid is $\underset{\_}{356.896K}$.

Explanation of Solution

The given data for the vapour pressure of nitric acid varies with temperature is written below as,

$\theta {/}^{°}\text{C}$	$0$	20	$40$	$50$	$70$	$80$	$90$	$100$
$p/\text{kPa}$	$1.92$	$6.38$	$17.7$	$27.7$	$62.3$	$89.3$	$124.9$	$170.9$

The normal boiling point is calculated by drawing the curve of $\ln p$ against $\frac{1}{T}$.  The data for the curve is shown below.

$\theta {/}^{°}\text{C}$	$p/\text{kPa}$	$\theta /\text{K}$	$\ln p$	$\frac{1}{T}/{\text{K}}^{-1}$
$0$	$1.92$	$273.15$	$0.652$	$0.00366$
$20$	$6.38$	$293.15$	$1.853$	$0.00341$
$40$	$17.7$	$313.15$	$2.873$	$0.00319$
$50$	$27.7$	$323.15$	$3.321$	$0.00309$
$70$	$62.3$	$343.15$	$4.131$	$0.00291$
$80$	$89.3$	$353.15$	$4.492$	$0.00283$
$90$	$124.9$	$363.15$	$4.827$	$0.00275$
$100$	$170.9$	$373.15$	$5.141$	$5.141$

The curve of $\ln p$ against $\frac{1}{T}$ from the above table is shown below.

The equation of the curve of $\ln p$ against $\frac{1}{T}$ is shown below.

    $y=-4569x+17.42$                                                                                        (1)

Compare the equation (1) with the straight line equation, $y=mx+C$.

The slope of the curve is $-4569$.

The intercept if the curve $\left(C\right)$ is $17.42$.

From the Clausius-Clapeyron’s equation, the slope of the curve of $\ln p$ against $\frac{1}{T}$ is equal to $\frac{-\Delta H_{\text{vaporization}}}{R}$.

Where,

• $\Delta H_{\text{vaporization}}$ is the enthalpy of vaporization.
• $R$ is the gas constant.

Therefore,

  $\frac{-\Delta H_{\text{vaporization}}}{R}=-4569\text{K}$                                                                                (2)

The value of the gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of $R$ in equation (2).

    $\begin{array}{c}\frac{-\Delta H_{\text{vaporization}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}=-4569\text{K}\\ -\Delta H_{\text{vaporization}}=-4569\text{K}\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =37986.66\text{J}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the enthalpy of vaporization of nitric acid is $37986.66\text{J}{\text{mol}}^{\text{-1}}$.

The normal boiling is calculated by the following formula.

    $\ln p=\frac{-\Delta H_{\text{vaporization}}}{RT}+C$                                                                                 (3)

Where,

• $p$ is the pressure at the normal boiling point.
• $R$ is the gas constant.
• $T$ is the normal boiling point.
• $C$ is the intercept of curve of $\ln p$ against $\frac{1}{T}$.
• $\Delta H_{\text{vaporization}}$ is the enthalpy of vaporization.

The normal boiling point of the nitric acid is calculated at $1\text{atm}$ pressure.

The conversion of $\text{atm}$ to $\text{kPa}$ is done as,

    $1\text{atm}=\text{101}\text{.325}\text{kPa}$

Therefore, the pressure at the normal boiling point is $\text{101}\text{.325}\text{kPa}$.

The value of the gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

The enthalpy of vaporization of nitric acid is $37986.66\text{J}{\text{mol}}^{\text{-1}}$.

The intercept of curve of $\ln p$ against $\frac{1}{T}$ from the above curve is $17.42$.

Substitute the values of $p,R,\Delta H_{\text{vaporization}}$ and the intercept $C$ in equation (3).

    $\begin{array}{c}\ln \left(101.325\right)=\frac{37986.66\text{J}{\text{mol}}^{\text{-1}}\text{}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times T}+17.42\\ 4.618=\frac{4568.99\text{K}}{T}+17.42\end{array}$

On further solving the above equation,

    $\begin{array}{c}4.618=\frac{4568.99\text{}\text{K}}{T}+17.42\\ \frac{4568.99\text{K}}{T}=17.42-4.618\\ T=\frac{4568.99\text{K}}{12.802}\\ =\underset{\_}{356.896K}\end{array}$

Hence, the normal boiling point of nitric acid is $\underset{\_}{356.896K}$.

(b)

Interpretation Introduction

Interpretation: The enthalpy of vaporization of nitric acid has to be calculated.

Concept introduction: The enthalpy of vaporization is calculated by using the Clausius-Clapeyron’s equation.  The Clausius-Clapeyron’s equation shows the correlation between temperatures and pressures for the phase transition process.

Answer to Problem 4B.9P

The enthalpy of vaporization of nitric acid is $\underset{\_}{37986.66Jmo{l}^{-1}}$.

Explanation of Solution

The given data for the vapour pressure of nitric acid varies with temperature is written below as,

$\theta {/}^{°}\text{C}$	$0$	20	$40$	$50$	$70$	$80$	$90$	$100$
$p/\text{kPa}$	$1.92$	$6.38$	$17.7$	$27.7$	$62.3$	$89.3$	$124.9$	$170.9$

The normal boiling point is calculated by drawing the curve of $\ln p$ against $\frac{1}{T}$.  The data for the curve is shown below.

$\theta {/}^{°}\text{C}$	$p/\text{kPa}$	$\theta /\text{K}$	$\ln p$	$\frac{1}{T}/{\text{K}}^{-1}$
$0$	$1.92$	$273.15$	$0.652$	$0.00366$
$20$	$6.38$	$293.15$	$1.853$	$0.00341$
$40$	$17.7$	$313.15$	$2.873$	$0.00319$
$50$	$27.7$	$323.15$	$3.321$	$0.00309$
$70$	$62.3$	$343.15$	$4.131$	$0.00291$
$80$	$89.3$	$353.15$	$4.492$	$0.00283$
$90$	$124.9$	$363.15$	$4.827$	$0.00275$
$100$	$170.9$	$373.15$	$5.141$	$5.141$

The curve of $\ln p$ against $\frac{1}{T}$ from the above table is shown below.

From the Clausius-Clapeyron’s equation the slope of the curve of $\ln p$ against $\frac{1}{T}$ is equal to $\frac{-\Delta H_{\text{vaporization}}}{R}$.

The slope of the curve of $\ln p$ against $\frac{1}{T}$  for the given data is $-4569$.

Therefore,

  $\frac{-\Delta H_{\text{vaporization}}}{R}=-4569\text{K}$                                                                                (4)

The value of the gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of $R$ in equation (4).

    $\begin{array}{c}\frac{-\Delta H_{\text{vaporization}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}=-4569\text{K}\\ -\Delta H_{\text{vaporization}}=-4569\text{K}\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{37986.66Jmo{l}^{-1}}\end{array}$

Hence, the enthalpy of vaporization of nitric acid is $\underset{\_}{37986.66Jmo{l}^{-1}}$.