Question

# $$\Delta H^o_\textit{f}$$ for $$NF_3 - 113\ kJ/mol$$. Bond energy for $$N - F$$ bond is $$27.5\ kJ/mo$$l. Calculate the bond energies of $$N_2$$ and $$F_2$$ if their magnitudes are in the ratio $$6 : 1$$ are, respectively

A
822.5,137.1 kJ/mol
B
979.8,163.3 kJ/mol
C
949.32,157.22 kJ/mol
D
762.6,127.1 kJ/mol

Solution

## The correct option is C $$949.32, 157.22\ kJ/mol$$$$\dfrac{1}{2} N_2 + \dfrac{3}{2} F_2 \rightarrow Nf_3$$$$\Delta H_f^o = -113 \, kJ/ mol$$  Bond                   Bond dissociation enthalapy$$N-F$$                      $$\Delta H_1 = 27.5$$$$N - N$$                     $$\Delta H_2 = 6x$$$$F - F$$                      $$\Delta H_3 = x$$$$275 \times 3 + (-113) = 6x \times \dfrac{1}{2} + \dfrac{3}{2} x$$$$712 = \dfrac{9}{2} x$$$$1424 = 9x$$$$x = \dfrac{1424}{9} = 158.22 kJ\, mol^{-1}$$Bond dissociation enthalapy of $$N_2 = 6 \times 158.22$$                                                           $$= 949.33 \, kJ \, mol^{-1}$$Bond dissociation enthalapy of $$F_2 = 158.22 \, kJ mol^{-1}$$option CChemistry

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