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 reacts, \(1196 \mathrm{~kJ}\) of heat is evolved. (a) Write a thermochemical equation for the reaction. (b) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{ClF}_{3} ?\)

First, we need to write the balanced chemical equation for the reaction of chlorine trifluoride with ammonia. The reaction can be written as: \(\mathrm{ClF}_{3} + \mathrm{NH}_{3} \rightarrow \mathrm{N}_{2} + \frac{3}{2}\mathrm{Cl}_{2} + 3\mathrm{HF}\) We need to balance the equation: 2\(\mathrm{ClF}_{3} + 6\mathrm{NH}_{3} \rightarrow 4\mathrm{N}_{2} + 3\mathrm{Cl}_{2} + 6\mathrm{HF}\) Now, we can include the heat evolved in the reaction. The problem states that when two moles of chlorine trifluoride reacts, \(1196 \mathrm{~kJ}\) of heat is evolved. Therefore, the thermochemical equation becomes: 2\(\mathrm{ClF}_{3} + 6\mathrm{NH}_{3} \rightarrow 4\mathrm{N}_{2} + 3\mathrm{Cl}_{2} + 6\mathrm{HF} + 1196\mathrm{~kJ}\)

To calculate the standard enthalpy of formation for chlorine trifluoride, we can use the given heat evolved for 2 moles of chlorine trifluoride in the reaction: \(1196 \mathrm{~kJ}\) evolved for 2 moles of \(\mathrm{ClF}_{3}\) We need to find the heat evolved for 1 mole of \(\mathrm{ClF}_{3}\): \(\frac{1196 \mathrm{~kJ}}{2} = 598\mathrm{~kJ}\) evolved for 1 mole of \(\mathrm{ClF}_{3}\) Since heat is evolved in the reaction, the enthalpy change, \(\Delta H\), is negative. Thus, the standard enthalpy of formation for \(\mathrm{ClF}_{3}\), denoted by \(\Delta H_{\mathrm{f}}^{\circ}\), is: \(\Delta H_{\mathrm{f}}^{\circ} (\mathrm{ClF}_{3}) = -598 \mathrm{~kJ/mol}\)