To calculate the energy change associated with a reaction, we can use the concept of bond energies. In this case, we need to determine the strength of the N-Br bond.
The given reaction involves breaking one N-Br bond in NBr3 and forming three new HO-Br bonds in HOBr. We also need to account for the bond energy of the existing N-H and O-H bonds.
The energy change (∆H) for the reaction can be calculated using the equation:
∆H = (sum of bond energies of the bonds broken) - (sum of bond energies of the bonds formed)
Analyzing the reaction:
Reactants:
1 N-Br bond (broken)
3 H-O bonds (broken)
Products:
3 H-O bonds (formed)
1 N-H bond (formed)
3 O-Br bonds (formed)
To calculate the energy change, we'll use the given bond energy values:
N-Br bond energy = 201 kJ/mol
O-H bond energy = 459 kJ/mol
N-H bond energy = 386 kJ/mol
O-Br bond energy = 201 kJ/mol
Let's now calculate the energy change (∆H) for the reaction:
∆H = (1 × N-Br bond energy broken) - (3 × O-H bond energy broken) + (3 × H-O bond energy formed) + (1 × N-H bond energy formed) + (3 × O-Br bond energy formed)
∆H = (1 × 201 kJ/mol) - (3 × 459 kJ/mol) + (3 × 459 kJ/mol) + (1 × 386 kJ/mol) + (3 × 201 kJ/mol)
∆H = 201 kJ/mol - 1377 kJ/mol + 1377 kJ/mol + 386 kJ/mol + 603 kJ/mol
∆H = 81 kJ/mol
Therefore, using the given bond energy values and the calculation, the energy change (∆H) for the reaction is +81 kJ/mol.
Now, to answer your specific question, the strength of the N-Br bond is given by the N-Br bond energy, which is 201 kJ/mol.