CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Enthalpy

Calculate the...

Question

Calculate the hydration enthalpy of ions for $N a C l (s)$ from the following data.
$Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}$
$Δ_{s o l} H = + 4 k J m o l^{- 1}$

A

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

- 1084 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

+ 1084 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 784 k J m o l^{- 1}$
We know that,
$Δ_{s o l n} H = Δ_{h y d} H + Δ_{l a t t i c e} H$
but given that,
$Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}$
$Δ_{s o l n} H = + 4 k J m o l^{- 1}$

putting the above values in the formula we get,
$+ 4 k J m o l^{- 1} = Δ_{h y d} H + 788 k J m o l^{- 1}$
$Δ_{h y d} H = - 784 k J m o l^{- 1}$

Theory:
Lattice enthalpy $Δ_{L a t t i c e} H$ :
Energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions.

Standard lattice enthalpy $Δ_{L a t t i c e} H^{o}$ :
Energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions at unit pressure $(1 b a r)$ and temperature of $298 K$ .

Standard enthalpy of hydration $Δ_{h y d} H^{o}$ Conditions specified = $1 b a r$ and $298 K$
Enthalpy change when one mole of anhydrous or partially hydrated salt combines with the requisite amount of water to form a new hydrated stable salt at unit pressure $(1 b a r)$ and $298 K$ .

Integral enthalpy of solution/enthalpy of dissolution:
Enthalpy change when one mole of the solute is dissolved in a definite amount of the solvent to make a solution.
$Δ_{s o l n} H = Δ_{h y d} H + Δ_{l a t t i c e} H$

Suggest Corrections

thumbs-up

0

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. 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 enthalpy of solid

N a C l

772 k J m o l^{- 1}

and of solution is

2 k J m o l^{- 1}

. If the hydration enthalpy of

N a^{+}

C l^{-}

ions are in the ratio of

3 : 2.5

, then the enthalpy of hydration of chloride ion is:

Q. Calculate the lattice enthalpy of

K C l

from the following data at standard states-
Enthalpy of sublimation of

K = 89 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} = 244 k J m o l^{- 1}

Ionisation energy of

K = 425 k J m o l^{- 1}

Electron gain enthalpy of

C l = - 355 k J m o l^{- 1}

Enthalpy of formation of

K C l = - 438 k J m o l^{- 1}

Q. The enthalpy of neutralization of a strong acid by a strong base is

- 57.32 {kJ mol}^{- 1}

. The enthalpy of formation of water and hydrogen ions are

- 285.84 {kJ mol}^{- 1}

0 {kJ mol}^{- 1}

respectively. The enthalpy of formation of the hydroxyl ion is:

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Standard XIII Chemistry

Join BYJU'S Learning Program

CrossIcon