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?H2(g) + Br2(g)→2HBr(g) Given that bond energy of H2, Br2 and HBr is 4.35 kJ mol-1,192 kJ mol-1 and 368 kJ mol -1 respectively.

To determine the enthalpy change (ΔHΔH) for the reaction H2(g)+Br2(g)→2HBr(g)H2(g)+Br2(g)→2HBr(g), we need to calculate the energy required to break all the bonds in the reactant molecules and subtract the energy required to form all the bonds in the product molecules.

Given bond energies:

• Bond energy of H2H2 (H-H) = 4.35 kJ/mol
• Bond energy of Br2Br2 (Br-Br) = 192 kJ/mol
• Bond energy of HBrHBr (H-Br) = 368 kJ/mol

The net change in enthalpy (ΔHΔH) can be calculated using the following formula:

ΔH=∑(bond energies of reactants)−∑(bond energies of products)ΔH=∑(bond energies of reactants)−∑(bond energies of products)

For the given reaction:

• Reactants: H2(g)+Br2(g)H2(g)+Br2(g)
• Products: 2HBr(g)2HBr(g)

Let's calculate:

1. Energy required to break bonds in reactants: H2(g)H2(g): 1 mole of H2H2 contains 1 bond, so energy required = 1×Bond energy of H21×Bond energy of H2 Br2(g)Br2(g): 1 mole of Br2Br2 contains 1 bond, so energy required = 1×Bond energy of Br21×Bond energy of Br2

   Total energy required to break bonds in reactants = Bond energy of H2+Bond energy of Br2Bond energy of H2+Bond energy of Br2

2. Energy released when bonds are formed in products: 2HBr(g)2HBr(g): 1 mole of 2HBr2HBr contains 2 bonds, so energy released = 2×Bond energy of HBr2×Bond energy of HBr

   Total energy released when bonds are formed in products = 2×Bond energy of HBr2×Bond energy of HBr

Now, let's substitute the given bond energies and calculate ΔHΔH:

ΔH=(Bond energy of H2+Bond energy of Br2)−(2×Bond energy of HBr)ΔH=(Bond energy of H2+Bond energy of Br2)−(2×Bond energy of HBr)

ΔH=(4.35 kJ/mol+192 kJ/mol)−(2×368 kJ/mol)ΔH=(4.35 kJ/mol+192 kJ/mol)−(2×368 kJ/mol)

ΔH=(196.35 kJ/mol)−(736 kJ/mol)ΔH=(196.35 kJ/mol)−(736 kJ/mol)

ΔH=−539.65 kJ/molΔH=−539.65 kJ/mol

Therefore, the enthalpy change (ΔHΔH) for the given reaction is -539.65 kJ/mol.