Question

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ᴛʜᴇ ʟᴀᴛᴛɪᴄᴇ ᴇɴᴛʜᴀʟᴘʏ ᴏꜰ ᴀɴ ɪᴏɴɪᴄ ᴄᴏᴍᴘᴏᴜɴᴅ ɪꜱ ᴛʜᴇ ᴇɴᴛʜᴀʟᴘʏ ᴡʜᴇɴ ᴏɴᴇ ᴍᴏʟᴇ ᴏꜰ ᴀɴ ɪᴏɴɪᴄ ᴄᴏᴍᴘᴏᴜɴᴅ ᴘʀᴇꜱᴇɴᴛ ɪɴ ɪᴛꜱ ɢᴀꜱᴇᴏᴜꜱ ꜱᴛᴀᴛᴇ, ᴅɪꜱꜱᴏᴄɪᴀᴛᴇꜱ ɪɴᴛᴏ ɪᴛꜱ ɪᴏɴꜱ. ɪᴛ ɪꜱ ɪᴍᴘᴏꜱꜱɪʙʟᴇ ᴛᴏ ᴅᴇᴛᴇʀᴍɪɴᴇ ɪᴛ ᴅɪʀᴇᴄᴛʟʏ ʙʏ ᴇxᴘᴇʀɪᴍᴇɴᴛ. ꜱᴜɢɢᴇꜱᴛ ᴀɴᴅ ᴇxᴘʟᴀɪɴ ᴀɴ ɪɴᴅɪʀᴇᴄᴛ ᴍᴇᴛʜᴏᴅ ᴛᴏ ᴍᴇᴀꜱᴜʀᴇ ʟᴀᴛᴛɪᴄᴇ ᴇɴᴛʜᴀʟᴘʏ ᴏꜰ ɴᴀᴄʟ(ꜱ).​

Answer 1

Answer:

The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. For the reaction

Na+Cl−(S)→Na+(g)+Cl−(g),ΔlatticeHΘ=+788kJmol−1

Since, it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber cycle.

Let us now calculate the enthalpy of Na+Cl− (s) by following steps given below.

(i) Na+(s)→Na(g), Sublimation of sodium metal, Δ⊂HΘ=108.4kJmol−1

(ii) Na(g)→Na+(g)+e−(g), The ionisation of sodium atoms, ionisation enthalpy ΔiHΘ=496kJmol−1

(iii) 12Cl2(g)→Cl(g), The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy 12ΔbondHΘ=121kJmol−1

(iv) Cl(g)+e−(g)→Cl−(g), electron gained by chlorine atoms. The electron gain enthalpy, ΔegHΘ=−348.6kJmol−1

(v) Na+(g)+Cl−(g)→Na+Cl−(s)

The sequence of steps is shown in given figure and is shown as Born-Haber cycle. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.

Applying Hess,s law, we get

ΔlatticeHΘ=411.2+108.4+121+496−348.6

ΔlatticeHΘ=+788kJ

Explanation:

hope it's work,!!!