Chapter 15, Problem 15.3P

Interpretation Introduction

Interpretation:

The heat energy required to change liquid mercury to solid has to be determined.

Concept Introduction:

Molar enthalpy for a process is the quantity of heat energy needed by one mole of a substance to undergo change in the process. The heat energy needed to change the temperature of $n$ moles of substance from $T_1$ to $T_2$ is as follows:

  $q=nC\left(T_1-T_2\right)$

Here,

$q$ is the heat energy in $\text{kJ}$.

$n$ is of moles of substance.

$C$ is specific heat capacity of substance.

$T_1$ and $T_2$ are initial and final temperatures of substance respectively.

Answer to Problem 15.3P

The heat energy needed to transform liquid mercury to solid is $-0.17908\text{kJ}$.

Explanation of Solution

The formula to determine moles of mercury is as follows:

  $\text{number}\text{of}\text{moles}\text{of}\text{mercury}=\frac{\text{mass}\text{of}\text{mercury}}{\text{molecular}\text{weight}\text{of}\text{mercury}}$        (1)

Substitute $200.59\text{g/mol}$ for molecular weight of mercury and $20.1\text{g}$ for mass of mercury in equation (1).

  $\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}\text{mercury}=\frac{20.1\text{g}}{200.59\text{g/mol}}\\ =0.1002\text{mol}\end{array}$

The value of $T_1$ is $25°\text{C}$.

The melting point of mercury is converted to Celsius as follows:

  $\begin{array}{c}{T}_2\left(°\text{C}\right)=T_2\left(\text{K}\right)-273.15\\ =234.32\text{K}-273.15\\ =-38.83°\text{C}\end{array}$

The value of $T_2$ is $-38.83°\text{C}$.

The temperature difference $\left(\Delta T\right)$ between $T_2$ and $T_1$ is calculated as follows:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =-38.83°\text{C}-25°\text{C}\\ =-63.83°\text{C}\end{array}$

The size difference on kelvin and Celsius scale is equal so $\left(\Delta T\right)$ is equal to $-63.83\text{K}$.

The formula to determine heat energy needed to change $20.1\text{g}$ of liquid mercury form $25°\text{C}$ to $-38.83°\text{C}$ is as follows:

  $q=nC\left(\Delta \text{T}\right)$        (2)

Substitute $28\text{J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$ for $C$, $0.1002\text{mol}$ for $n$ and $-63.83\text{K}$ for $\Delta T$ in equation (2) to calculate heat energy needed to transform $20.1\text{g}$ of liquid mercury form $25°\text{C}$ to $-38.83°\text{C}$.

  $\begin{array}{c}q=\left(0.1002\text{mol}\right)\left(28\text{J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\left(-63.83\text{K}\right)\\ =\left(2.8056\times {10}^{-3}\text{kJ}\cdot {\text{K}}^{-1}\right)\left(-63.83\text{K}\right)\\ =-0.17908\text{kJ}\end{array}$

The formula to calculate the heat energy that must be released from the liquid mercury to fuse it into solid is as follows:

  $q_{\text{fus}}=n\left(-\Delta H_{\text{fus}}\right)$        (3)

Substitute $2.29\text{kJ}\cdot {\text{mol}}^{-\text{1}}$ for $\Delta H_{\text{fus}}$ and $0.1002\text{mol}$ for $n$ in equation (3) to calculate the heat energy needed to fuse liquid mercury to solid at melting point.

  $\begin{array}{c}{q}_{\text{fus}}=\left(0.1002\text{mol}\right)\left(-2.29\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\\ =-0.2294\text{kJ}\end{array}$

The total energy that must be removed from $20.1\text{g}$ of mercury is calculated as follows:

  $\begin{array}{c}\text{total}\text{energy}=q+q_{\text{fus}}\\ =\left(-\text{0}\text{.17908}\text{kJ}\right)+\left(-\text{0}\text{.2294}\text{kJ}\right)\\ =-\text{0}\text{.40848}\text{kJ}\end{array}$

The total energy that must be removed from $20.1\text{g}$ of mercury to convert it into liquid to solid is $-\text{0}\text{.40848}\text{kJ}$.