If the bond energies of $ ext{H}- ext{H}$, $ ext{Br}- ext{Br}$, and $ ext{H}- ext{Br}$ are $433$, $192$,

Question

If the bond energies of $	ext{H}-	ext{H}$, $	ext{Br}-	ext{Br}$, and $	ext{H}-	ext{Br}$ are $433$, $192$, and $364	ext{ kJ mol}^{-1}$ respectively, the $\Delta H^\circ$ for the reaction $	ext{H}_2(g) + 	ext{Br}_2(g) ightarrow 2	ext{HBr}(g)$ will be:

Accepted Answer

The standard enthalpy of reaction ($\Delta_r H^\circ$) can be calculated from the bond energies of the reactants and products using the formula:
$\Delta_r H^\circ = \sum 	ext{Bond energies of reactants} - \sum 	ext{Bond energies of products}$
For the reaction: $	ext{H}_2(g) + 	ext{Br}_2(g) ightarrow 2	ext{HBr}(g)$
$\Delta_r H^\circ = [	ext{BE}(	ext{H}-	ext{H}) + 	ext{BE}(	ext{Br}-	ext{Br})] - [2 	imes 	ext{BE}(	ext{H}-	ext{Br})]$
Given values:
$	ext{BE}(	ext{H}-	ext{H}) = 433	ext{ kJ mol}^{-1}$
$	ext{BE}(	ext{Br}-	ext{Br}) = 192	ext{ kJ mol}^{-1}$
$	ext{BE}(	ext{H}-	ext{Br}) = 364	ext{ kJ mol}^{-1}$
Substituting the values:
$\Delta_r H^\circ = [433 + 192] - [2 	imes 364]$
$\Delta_r H^\circ = 625 - 728$
$\Delta_r H^\circ = -103	ext{ kJ}$

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