Question

The standard enthalpy of formation of $$N{H}_3$$ is 46.0 kJ/mol. If the enthalpy of formation of $$H_2$$ from its atoms is -436 kJ/mol and that of $$N_2$$ is -712 kJ/mol, the average bond enthalpy of N — H bond in $$N{H}_3$$ is:

a) -1102 kJ/mol

b) -964 kJ/mol

c) +352 kJ/mol

d) +1056 kJ/mol

Answer 1

Hint: Bond dissociation enthalpy/energy is defined as the amount of energy required to dissociate a particular bond present in a compound. It is done via homolytic fracture which produces the radical species as a result.

$$N_2+BondEnergy\to 2N$$

Complete step by step solution:

The chemical reaction for the formation of Ammonia ($$N{H}_3$$) can be given as:

$${\displaystyle \frac{1}{2}}{N}_2+{\displaystyle \frac{3}{2}}{H}_2\to N{H}_3$$

We have been given in the question that the:

Standard enthalpy of formation of $$N{H}_3$$ ($${\mathrm{\Delta }}_{f}H$$) = 46.0 kJ/mol.

Enthalpy of formation or Bond enthalpy of $$H_2$$ from its atoms ($$\mathrm{\Delta }{H}_{H_2}$$)= -436 kJ/mol

Enthalpy of formation or Bond enthalpy of $$N_2$$ from its atoms ($$\mathrm{\Delta }{H}_{N_2}$$) = -712 kJ/mol

We know that,

$$\mathrm{\Delta }{\text{H}}_{\text{rxn}}\text{=}\sum \text{Bond enthalpy of}\text{product-}\sum \text{Bond enthalpy}\text{of}\text{reactants}$$

$${\mathrm{\Delta }}_{f}H=[{\displaystyle \frac{1}{2}}\mathrm{\Delta }{H}_{N_2}+{\displaystyle \frac{3}{2}}\mathrm{\Delta }{H}_{H_2}]-3\left[\mathrm{\Delta }{H}_{N-H}\right]$$

Putting the values in above equation, we get,

$$-46=[{\displaystyle \frac{1}{2}}\times 712+{\displaystyle \frac{3}{2}}\times 436]-3\left[\mathrm{\Delta }{H}_{N-H}\right]$$

After rearranging,

$$3\left[\mathrm{\Delta }{H}_{N-H}\right]=1010+46$$

$$\left[\mathrm{\Delta }{H}_{N-H}\right]={\displaystyle \frac{1053}{3}}$$

$$\left[\mathrm{\Delta }{H}_{N-H}\right]=+352kJmo{l}^{-1}$$

Therefore, the average bond enthalpy of N-H bond in NH${}_3$ is $$+352kJmo{l}^{-1}$$.

So, the correct option is (c).

Note: Enthalpy of formation or bond formation enthalpy is given in the question is negative but the reaction mechanism we have taken in the answer involves bond breaking. Hence, the values need to be taken as positive.