delta Hrxn = (sum bonds broken reactants) - (sum bonds formed products).
Solve for NBr bonds and divide by 3.
reaction
NBr3(g) + 3H2O(g) -> 3HOBr(g) + NH3(g)
is +81 kJ/mol rxn.
These bond energy values might be useful:
O-H 459 kJ/mol; N-H 386 kJ/mol; O-Br 201
kJ/mol.
The strength of the N Br bond is
1. 4 kJ/mol
2. 465 kJ/mol
3. 66 kJ/mol
4. 128 kJ/mol
5. 248 kJ/mol
6. 280 kJ/mol
7. 155 kJ/mol
Solve for NBr bonds and divide by 3.
Given bond energy values:
O-H bond energy = 459 kJ/mol
N-H bond energy = 386 kJ/mol
O-Br bond energy = 201 kJ/mol
In the reactants:
NBr3(g) has 1 N-Br bond and 3 N-H bonds.
3H2O(g) has 6 O-H bonds.
In the products:
3HOBr(g) has 3 O-Br bonds and 3 O-H bonds.
NH3(g) has 3 N-H bonds.
Now let's calculate the bond energy changes:
Reactants:
N-Br bond energy = 1 * (N-H bond energy) + 3 * (H2O bond energy)
= 1 * 386 kJ/mol + 3 * 459 kJ/mol
= 1613 kJ/mol
Products:
O-Br bond energy = 3 * (O-Br bond energy)
= 3 * 201 kJ/mol
= 603 kJ/mol
O-H bond energy = 3 * (O-H bond energy)
= 3 * 459 kJ/mol
= 1377 kJ/mol
N-H bond energy = 3 * (N-H bond energy)
= 3 * 386 kJ/mol
= 1158 kJ/mol
Total bond energy of products = O-Br bond energy + O-H bond energy + N-H bond energy
= 603 kJ/mol + 1377 kJ/mol + 1158 kJ/mol
= 3138 kJ/mol
Total bond energy of reactants = N-Br bond energy
= 1613 kJ/mol
Energy change (ΔH) = Total bond energy of products - Total bond energy of reactants
= 3138 kJ/mol - 1613 kJ/mol
= 1525 kJ/mol
The energy change associated with the reaction is +81 kJ/mol rxn, as given in the question.
To solve for the strength of the N-Br bond, we can rearrange the equation as follows:
N-Br bond energy = Total bond energy of reactants - Total bond energy of products
= 1613 kJ/mol - 3138 kJ/mol
= -1525 kJ/mol
Since bond energies are typically positive values, the negative sign indicates an exothermic reaction where energy is released. Therefore, the strength of the N-Br bond is:
N-Br bond energy = 1525 kJ/mol
None of the given options match this calculated value, so the correct answer is none of the listed options.
The given reaction equation is:
NBr3(g) + 3H2O(g) -> 3HOBr(g) + NH3(g)
To break the bonds on the reactant side, we need to consider the bonds in NBr3 and H2O. The bonds being broken are N-Br and O-H.
Bond energy of N-Br bond = 201 kJ/mol (given)
Bond energy of O-H bond = 459 kJ/mol (given)
To form the bonds on the product side, we need to consider the bonds in HOBr and NH3. The bonds being formed are O-Br and N-H.
Bond energy of O-Br bond = ?
Bond energy of N-H bond = ?
We can use the given bond energy values and the energy change (enthalpy change) of the reaction to calculate the bond energy of the O-Br bond.
Energy change (∆H) of the reaction = energy absorbed (bonds broken) - energy released (bonds formed)
Given: ∆H = +81 kJ/mol (enthalpy change)
The bonds broken in the reaction are N-Br and O-H. The bonds formed are O-Br and N-H.
Energy absorbed (bonds broken) = bond energy of N-Br (201 kJ/mol) + 3 × bond energy of O-H (459 kJ/mol)
Energy released (bonds formed) = 3 × bond energy of O-Br + bond energy of N-H
Substituting the given values:
Energy absorbed = 201 kJ/mol + 3 × 459 kJ/mol
Energy released = 3 × bond energy of O-Br + bond energy of N-H
Using the given information and rearranging the equation, we can find the bond energy of O-Br:
Bond energy of O-Br = (energy released - energy absorbed)/3
Let's calculate this value:
Energy absorbed = 201 kJ/mol + 3 × 459 kJ/mol = 1578 kJ/mol
Energy released = 81 kJ/mol (given)
Bond energy of O-Br = (81 kJ/mol - 1578 kJ/mol)/3
= -1497 kJ/mol/3
= -499 kJ/mol
The negative sign indicates that energy is released when the bond is formed.
So, the bond energy of O-Br is approximately 499 kJ/mol.
Unfortunately, we do not have the bond energy value for N-H, which is required to calculate the strength of the N-Br bond. Without this information, we cannot determine the strength of the N-Br bond.
Therefore, none of the options provided (1, 2, 3, 4, 5, 6, or 7) can be selected as the correct answer.