The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.

$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$

Give that bond energy of $H_{2}, B r_{2}$ and $H B r$ is $435 k J m o l^{- 1}, 192 k J m o l^{- 1}$ and $368 k J m o l^{- 1}$ respectively.

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Solution

$Net enthalpy change of reaction (Δ_{r} H)$

The enthalpy change of the given reaction can be calculated from the relation:

$Δ_{r} H = Σ$ bond energy (reactants) - $Σ$ bond energy (products)

For the reaction:

$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$

As we know, the net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules.

So, as per given reaction, net enthalpy change $(Δ_{r} H)$ would be

$Δ_{r} H =$ (Bond energy of $H - H$ bond + Bond energy of $B r - B r$ bond)) – ( $2 \times$ bond energy of $H - B r$ bond)

$= (435 + 192) k J m o l^{- 1} - (2 \times 368) k J m o l^{- 1}$

$Δ_{r} H = - 109 k J m o l^{- 1}$

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