Question

Chlorine trifluoride is a toxic, intensely reactive gas. It was used in World War II to make incendiary bombs. It reacts with ammonia and forms nitrogen, chlorine, and hydrogen fluoride gases. When two moles of chlorine trifluoride react, \(1196 \mathrm{~kJ}\) of heat are evolved. (a) Write a thermochemical equation for the reaction. (b) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{ClF}_{3}\) ?

Short answer

Based on the step-by-step analysis, the thermochemical equation for the given reaction is as follows: $$ 2\,\text{ClF}_3(\text{g}) + 6\,\text{NH}_3(\text{g}) \rightarrow 2\,\text{N}_2(\text{g}) + 3\,\text{Cl}_2(\text{g}) + 12\,\text{HF}(\text{g})\quad\Delta H=-1196\mathrm{kJ} $$ The standard enthalpy of formation (\(\Delta H_{\mathrm{f}}^{\circ}\)) for chlorine trifluoride is: $$ \Delta H_{\mathrm{f}}^{\circ}(\text{ClF}_3)=-598\,\frac{\mathrm{kJ}}{\text{mole}} $$

Step-by-step solution

Write a balanced chemical equation

First, let's write the balanced chemical equation for the reaction between chlorine trifluoride, ClF\(_3\), and ammonia, NH\(_3\), forming nitrogen, N\(_2\), chlorine, Cl\(_2\), and hydrogen fluoride, HF: $$ 2\,\text{ClF}_3(\text{g}) + 6\,\text{NH}_3(\text{g}) \rightarrow 2\,\text{N}_2(\text{g}) + 3\,\text{Cl}_2(\text{g}) + 12\,\text{HF}(\text{g}) $$ Here, we have balanced the chemical equation considering the moles of each reactant.

Write the thermochemical equation

Given that two moles of chlorine trifluoride evolve \(1196\mathrm{~kJ}\) of heat, we will include that value in the balanced chemical equation as the heat evolved or released, represented by \(\Delta H\). Here, the negative sign refers to the fact that heat is released: $$ 2\,\text{ClF}_3(\text{g}) + 6\,\text{NH}_3(\text{g}) \rightarrow 2\,\text{N}_2(\text{g}) + 3\,\text{Cl}_2(\text{g}) + 12\,\text{HF}(\text{g})\quad\Delta H=-1196\mathrm{~kJ} $$

Calculate the standard enthalpy of formation for chlorine trifluoride

To calculate the standard enthalpy of formation (\(\Delta H_{\mathrm{f}}^{\circ}\)) for \(\mathrm{ClF}_{3}\), we need to consider that the heat evolved (-1196 kJ) is for two moles of chlorine trifluoride. We will divide the total heat evolved by the number of moles (2 moles) of ClF\(_3\) to get the \(\Delta H_{\mathrm{f}}^{\circ}\) for one mole of ClF\(_3\): $$ \Delta H_{\mathrm{f}}^{\circ}(\text{ClF}_3)=\frac{-1196\mathrm{~kJ}}{2\,\text{moles}}=-598\,\frac{\mathrm{kJ}}{\text{mole}} $$ So, the standard enthalpy of formation for one mole of \(\mathrm{ClF}_{3}\) is \(-598\,\frac{\mathrm{kJ}}{\text{mole}}\).