Question

From the following data, calculate the average bond enthalpy for the NOH bond: NH3(g) ¡NH2(g) 1 H(g) ¢H° 5 435 kJ/mol NH2(g) ¡NH(g) 1 H(g) ¢H° 5 381 kJ/mol NH(g) ¡N(g) 1 H(g) ¢H° 5 360 kJ/mol

Answer 1

Answer:

The average bond enthalpy for the N-H bond is 392 kJ/mol.

Explanation:

$NH_3(g) \rightarrow NH_2(g) +1 H(g),\Delta H^o_{1}= 435 kJ/mol$..[1]

$NH_2(g) \rightarrow NH(g) + H(g),\Delta H^o_{2}= 381 kJ/mol$..[2]

$NH(g) \rightarrow N(g)+ 1 H(g),\Delta H^o_{3}= 360 kJ/mol$..[3]

On adding [1] , [2] and [3] , we get:

$NH_3(g) \rightarrow N(g) +3H(g),\Delta H^o_{4}= ?$

$\Delta H^o_{4}=\Delta H^o_{1}+\Delta H^o_{2}+\Delta H^o_{3}$

$= 435 kJ/mol+381 kJ/mol +360 kJ/mol = 1,176 kJ/mol$

There three bonds N-H bond sin ammonia. Then average bond enthalpy of single N-H bond is :

$\Delta H^o_{N-H}=\frac{1,176 kJ/mol}{3}=392 kJ/mol$

The average bond enthalpy for the N-H bond is 392 kJ/mol.