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  • find the enthaply of formation of BrCl is

Given:

\\ \Delta H_{vap} \textup{ of Bromine} = X\ kJMol^{-1} \\ \Delta_{a} H \ of\ Br_{2} (g) = Y\ kJMol^{-1} \\ \Delta_{a} H \ of\ Br_{2} (g) = Z\ kJMol^{-1} \\ \Delta_{sublimation} H \ of\ BrCl(s) = W\ kJMol^{-1} \\ \Delta_{a} H \ of\ BrCl(g) = a\ kJMol^{-1}

The Enthalpy of the formation of BrCl(s) is ?

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best_answer

\\Br_{(l)}+Cl_{(g)}\rightarrow BrCl_{(g)}\overset{Deposition}{\rightarrow} BrCl_{(s)} \\ 1) \frac{1}{2}(Br_{2}(l)\overset{Vap}{\rightarrow} Br_{2}(g)) \qquad \qquad \Delta H = \frac{x}{2} kJ/mol \\ 2) \frac{1}{2}(Br_{2}(l)\rightarrow Br_{2}(g)) \qquad \qquad \Delta H = \frac{y}{2} kJ/mol \\ 3) \frac{1}{2}(Cl_{2}(l)\rightarrow Br_{2}(g)) \qquad \qquad \Delta H = \frac{z}{2} kJ/mol \\ 4) (Br(g)+Cl(g)\rightarrow BrCl(g)) \qquad \Delta H = -a \ kJ/mol \\ 5) ( BrCl(g))\overset{deposition}{\rightarrow} BrCl(s)) \qquad \qquad \Delta H = -w \ kJ/mol \\ Final\Rightarrow \frac{1}{2}Br_2(l)+\frac{1}{2}Cl_2(g)\rightarrow BrCl(s) \qquad \Delta H = \frac{x}{2}+ \frac{y}{2}+ \frac{z}{2}-w-a \ kJ/mol

 

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I hope your doubt is cleared regarding the formation enthalpy of BrCl
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