The enthalpy change for the atomisation of 1010 molecules of ammonia i

1.	The enthalpy change for the atomisation of 1010 molecules of ammonia is 1.94 × 10-11 kJ.Calculate the bond enthalpy of N-H bond.
Answer» Given : Enthalpy change for atomisation of 10¹⁰ molecules = 1.94 × 10^-11kJ Number of NH₃ molecules dissociate = 10¹⁰ Bond enthalpy of N-H = ΔH = ? 1 mole of NH₃ contains 6.022 × 10²³NH₃ molecules. ∵ For atomisation of 1010 molecules of NH₃ ΔH = 1.94 × 10^-11kJ ∴ For atomisation of 6.022 × 10²³ molecules of NH₃, ΔH = \(\frac{1.94\times 10^{-11}\times 6.022\times 10^{23}}{10^{10}}\) = 1168 kJ mol^-1 In NH₃ three N-H bonds are broken on atomisation. NH_3(g) → N_(g)+ 3H_(g)ΔH = 1168 kJ mol^-1 ∴ Average bond enthalpy of N-H bond is, ΔH = \(\frac{1168}{3}\) = 389.3 kJ mol^-1 ∴ Bond enthalpy of N-H bond = 389.3 kJ mol^-1.

The enthalpy change for the atomisation of 1010 molecules of ammonia is 1.94 × 10-11 kJ.Calculate the bond enthalpy of N-H bond.

Answer»

Given :

Enthalpy change for atomisation of 10¹⁰

molecules = 1.94 × 10^-11kJ

Number of NH₃ molecules dissociate = 10¹⁰

Bond enthalpy of N-H = ΔH = ?

1 mole of NH₃ contains 6.022 × 10²³NH₃ molecules.

∵ For atomisation of 1010 molecules of NH₃

ΔH = 1.94 × 10^-11kJ

∴ For atomisation of 6.022 × 10²³ molecules of NH₃,

ΔH = \(\frac{1.94\times 10^{-11}\times 6.022\times 10^{23}}{10^{10}}\)

= 1168 kJ mol^-1

In NH₃ three N-H bonds are broken on atomisation.

NH_3(g) → N_(g)+ 3H_(g)ΔH = 1168 kJ mol^-1

∴ Average bond enthalpy of N-H bond is,

ΔH = \(\frac{1168}{3}\) = 389.3 kJ mol^-1

∴ Bond enthalpy of N-H bond = 389.3 kJ mol^-1.

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