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Lattice energy of NaCl((s)) is -788kJ mo...

Lattice energy of `NaCl_((s))` is `-788kJ mol^(-1)` and enthalpy of hydration is `-784kJ mol^(-1)`. Calculate the heat of solution of `NaCl_((s))`.

A

`4 kJ mol^(-1)`

B

`-4 kJ mol^(-1)`

C

`-1572 kJ mol^(-1)`

D

`1572 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`NaCl(s) rarr Na^(+) + Cl^(-)" "……(i)`
`Na^(+) + Cl^(-) + (aq) rarr NaCl(aq)" "…..(ii)`
Adding (i) and (ii) : `DeltaU = nC_(p) Delta T`
Heat of solution of `NaCl(s) = 788 + (-788 + (-784) = 4 kJ//mol`.
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