Question

Question: Use the following standard enthalpies of formation to estimate the NH bond energy in ammonia. Compare this with the value in Table 13.6.

N_(g) 472.7 KJ/mol

H_(g) 216.0 KJ/mol

NH3_(g) - 46.1 KJ/mol

Short answer

The reaction is as follows -

N_2(g) + 3H_{2 (g)}2NH_{3 (g)}

𝛥H_f ^o (N) - 472.7 KJ/mol

𝛥H_f ^o (H) - 216.0 KJ/mol

𝛥H_f ^o (NH₃) - - 46.1 KJ/mol

Step-by-step solution

Describing the bond energy of the NH bond

The enthalpy of reaction (𝛥H)

=𝞢𝛥H_(products) -𝞢𝛥H_(reactants)

By putting values,

𝛥H = 2 *𝛥H_f ^o (NH₃) - [2 *𝛥H_f ^o (N) + 6 *𝛥H_f ^o (H)]

= 2 * (- 46.1) - [2 * 472.7 + 6 * 216.0]

= -92.2 - [855.4 + 1296]

= -2243.6 KJ/mol

Therefore, the enthalpy of reaction (𝛥H) is-2243.6 KJ/mol.

Now,

𝛥H =𝞢BE_(reactants) –𝞢BE_(products)

Therefore,

𝛥H = [BE (NN) + 3 * BE (H –H)] - 6 * BE (N –H)

= [941 + 3 * 432] - 6 * BE (N –H)

= 941 +1296 – 6 * BE (N –H)

= 2237 – 6 * BE (N –H)

= 2237 – 6 * BE (N –H)

BE (N –H) = 4480.6 / 6

BE (N –H) = 746.67 KJ/mol

The bond energy of the N –H bond is746.67 KJ/mol.

 Comparing the values of bond energy between calculated and average 

The calculated bond energy of the N –H bond is 746.67 KJ/mol.

The average bond energy of the N –H bond is 391 KJ/mol.

Here, the calculated value of bond energy is high compared to the average value.