Question

Question: Use the reaction

$\mathbf{2}\mathit{C}\mathit{l}{\mathit{F}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathit{H}\mathit{F}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}{\mathit{H}}_{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{1196}\mathbf{}\mathit{k}\mathit{J}$

to$\mathbf{∆}{{\mathit{H}}_{\mathbf{f}}}^{\mathbf{0}}$ calculate for ClF₃(g).

Short answer

The value of$∆{H_f}^0$for ClF₃(g) is -169 kJ/mol

Step-by-step solution

Determine the enthalpy value

$\begin{array}{rcl}2Cl{F}_3\left(g\right)+2N{H}_3\left(g\right)& \to & N_2\left(g\right)+6HF\left(g\right)+C{l}_2\left(g\right)∆H_0=-1196kJ\end{array}$

$\begin{array}{rcl}∆{H^0}_{f,HF}& =& -271kJ/mol\\ ∆{H^0}_{f,N{H}_3}& =& -46kJ/mol\\ ∆H_0& =& \left(6\times ∆{H^0}_{f,HF}\right)-\left(2\times ∆{H^0}_{f,Cl{F}_3}+2\times ∆{H^0}_{f,N{H}_3}\right)\\ -1196kJ& =& \left(6\times -271kJ/mol\right)-2mol\times ∆{H^0}_{f,Cl{F}_3}-\left(2\times -46kJ/mol\right)\\ -1196kJ& =& -1626kJ-2mol\times ∆{H^0}_{f,Cl{F}_3}+92kJ\\ ∆{H^0}_{f,Cl{F}_3}& =& -169kJ/mol\\ & & \end{array}$