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Ionisation Energy

If Lattice en...

Question

If Lattice enthalpy and hydration enthalpy for 1 mole of NaCl are 788 KJmol^-1and-784 KJ mol^-1respectively then enthalpy of solution of NaCl(s)would be
(1)-4KJmol^-1
(2)+4KJmol^-1
(3)-8KJ mol^-1
(4)+8KJ mol^-1
Explain.

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Solution

H_hyd= H_lattice+ H_solv
-784 = 788 + H_solv
H_{solv =}-784 -(-788) = +4kJ

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Similar questions

Q. Calculate the hydration enthalpy of ions for

N a C l (s)

from the following data.

Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}

Δ_{s o l} H = + 4 k J m o l^{- 1}

Q. The change in enthalpy for hydration of ( i ) the chloride ion ; ( ii ) the iodide ion are :
Given :
* enthalpy change of solution of NaCl(s)

= - 2

kJ / mol.
* enthalpy change of solution of NaI(s)

= + 2

kJ / mol.
* enthalpy change of hydration of

N a^{+} (g) = - 390

kJ / mol.
* lattice enthalpy of NaCl

= - 772

kJ / mol.
* lattice enthalpy of NaI

= - 699

Q. Calculate the hydration enthalpy of ions for

N a C l (s)

from the following data.

Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}

Δ_{s o l} H = + 4 k J m o l^{- 1}

Q. The lattice energy of

N a C l (s)

- 790 k J . {m o l}^{- 1}

and enthalpy of hydration is

- 785 k J . {m o l}^{- 1}

. Calculate the enthalpy of the solution of

N a C l (s)

.

Q. The enthalpy of dissolution(or solution) of sodium chloride is

4 k J m o l^{- 1}

and its enthalpy of hydration of ions is

- 784 k J m o l^{- 1}

. What will be the lattice enthalpy of sodium chloride?

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