Answer :
Answer:
The average bond enthalpy for the N-H bond is 392 kJ/mol.
Explanation:
[tex]NH_3(g) \rightarrow NH_2(g) +1 H(g),\Delta H^o_{1}= 435 kJ/mol [/tex]..[1]
[tex]NH_2(g) \rightarrow NH(g) + H(g),\Delta H^o_{2}= 381 kJ/mol [/tex]..[2]
[tex]NH(g) \rightarrow N(g)+ 1 H(g),\Delta H^o_{3}= 360 kJ/mol [/tex]..[3]
On adding [1] , [2] and [3] , we get:
[tex]NH_3(g) \rightarrow N(g) +3H(g),\Delta H^o_{4}= ? [/tex]
[tex]\Delta H^o_{4}=\Delta H^o_{1}+\Delta H^o_{2}+\Delta H^o_{3}[/tex]
[tex] = 435 kJ/mol+381 kJ/mol +360 kJ/mol = 1,176 kJ/mol[/tex]
There three bonds N-H bond sin ammonia. Then average bond enthalpy of single N-H bond is :
[tex]\Delta H^o_{N-H}=\frac{1,176 kJ/mol}{3}=392 kJ/mol[/tex]
The average bond enthalpy for the N-H bond is 392 kJ/mol.
Based on the data provided, the average bond enthalpy for the N-H bond is 392 kJ/mol.
What is enthalpy change of a reaction?
The enthalpy change of a reaction is the energy released or absorbed when reactant molecules react to form products.
The data of the entahpies of the stepwise reaction for the formation of NH3 is given below:
- NH3(g) → NH2(g) + H(g) ΔH° = 435 kJ/mol
- NH2 (g) → NH (g) + H(g) ΔH° = 381 kJ/mol
- NH (g) → N (g) + H(g) ΔH° = 360 kJ/mol
On adding [1] , [2] and [3];
NH3 (g)→N (g) + 3H(g) ΔH° = 453 + 381 + 360
ΔH° = 1,176 kJ/mol
Since there are three bonds N-H bonds in NH3, therefore, average bond enthalpy of single N-H bond is given as:
Average bond enthalpy for the N-H bond = 1,176 kJ/mol / 3
Average bond enthalpy for the N-H bond = 392kJ/mol
Therefore, the average bond enthalpy for the N-H bond is 392 kJ/mol.
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