Question

Question: Suppose $\mathbf{32}\mathbf{.}\mathbf{1}\mathbf{}{\mathbf{gClF}}_{\mathbf{3}}\mathbf{(}\mathbf{}\mathbf{g}\mathbf{)}$ and$\mathbf{17}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{g}\mathbf{}\mathbf{Li}\mathbf{(}\mathbf{s}\mathbf{)}$ are mixed and allowed to react at atmospheric pressure and ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$until one of the reactants is used up, producing$\mathbf{LiCl}\mathbf{(}\mathit{s}\mathbf{)}$ and$\mathbf{LiF}\mathbf{(}\mathit{s}\mathbf{)}$ . Calculate the amount of heat evolved.

Short answer

The amount of heat evolved is 732.66kJ.

Step-by-step solution

Given data

Mass of each component is given as:

$\begin{array}{l}m\left({\mathrm{ClF}}_3\right)=32.1\mathrm{g}\\ m\left(\mathrm{Li}\right)=17.3\mathrm{g}\end{array}$

Standard enthalpies of formation for each component are given as:

$\begin{array}{c}\Delta H_f\left({\mathrm{ClF}}_3,g\right)=-163.2\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f(\mathrm{Li},s)=0\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f(\mathrm{LiCl},s)=-408.61\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f(\mathrm{LiF},s)=-615.91\mathrm{kJ}/\mathrm{mol}\end{array}$

Concept of heat evolved in reaction

The amount of heat evolved during a combustion process by calculating the enthalpy change of formation for every molecule involved in the chemical reaction for one mole by given formula.

$\mathit{\Delta }\mathit{H}\mathbf{=}\mathit{n}\mathbf{\times }\mathit{\Delta }{\mathit{H}}_{\mathbf{m}}$

Where, $n$is number of total moles and $\Delta H_m$is change in enthalpy

Calculation of enthalpy change

The formula is, $\Delta H=\sum \Delta H_{f_{\mathrm{products}}}-\sum \Delta H_{f_{\mathrm{reactants}}}$.

For this reaction, the change in enthalpy would be:

$\begin{array}{l}\Delta H_m=\Delta H_f(\mathrm{LiCl},s)+3\times \Delta H_f(\mathrm{LiF},s)-\Delta H_f\left({\mathrm{ClF}}_3,g\right)-4\times \Delta H_f(\mathrm{LiF},s)\\ \Delta H_m=-408.61+3\times (-615.97)-(-163.2)-4\times 0\\ \Delta H_m=-2093.32\mathrm{kJ}/\mathrm{mol}\end{array}$

Determine limit reactant

The molar mass of reactants is given below.

$\begin{array}{c}M\left({\mathrm{ClF}}_3\right)=92.448\mathrm{g}/\mathrm{mol}\\ M\left(\mathrm{Li}\right)=6.941\mathrm{g}/\mathrm{mol}\end{array}$

For determination of limiting reactant find the number of moles of each reactant.

$\begin{array}{c}n=\frac{m}{M}\\ \mathrm{n}\left({\mathrm{ClF}}_3\right)=\frac{32.1}{92.448}\\ =0.35\mathrm{mol}\\ n\left(Li\right)=\frac{17.3}{6.941}\end{array}$

So, $n\left(Li\right)=2.5\mathrm{m}ol$

From the reaction, four moles of $Li$ required for one mol of $CI{F}_3$.

So for $0.35mol$of $CI{F}_3$.

$0.35\times 4=1.4\mathrm{mol}$of $Li$required.

This means that $CI{F}_3$ is the limiting reactant.

Calculate heat evolved

The heat is calculated by given equation.

$\begin{array}{l}\Delta H=n\times \Delta H_m\\ \Delta H=0.35\times 2093.32\\ \Delta H=732.66\mathrm{kJ}\end{array}$