Question

calculate the lattice enthalpy of sodium chloride using born haber cycle

Answer 1

Answer:

written below-

Explanation:

Na(s)+φ Cl2 △Hf >NaCl(s) △H1 △H3 |uCl(g) cr(g) △H4 +Na(g) Na^{-} (g) △H2 △Hf= heat of formation of sodium chloride =-411.3 kJ mol^{-1} △H1= heat of sublimation of N a(g)=108. kJ mol^{-l} △H2= ionisation energy of I Ja(g)=495.0 kJ mol^{-1} △H3= dissociation energy of C l2(s)=244 k J mol- △H4= Electron affinity of Cl(s)=-349.0 kJ mol-l U= lattice energy of NaCl △Hf=△H1+△H2++△H3+△H4+U ∴ U=(△Hf)-(△H1+△H2++ 2△H3+△H4) = >U=(-411.3)-(108.7+495+122-349) U=(-411.3)-(376.7) ∴ U=-788 kJ mol^{-1} This negative sign in lattice energy indicates that the energy is released when sodium is formfrom its constituent gaseous ions Na^{+} and Cl`

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