Home
Class 11
CHEMISTRY
The lattice energy of NaCl is +788 kJ mo...

The lattice energy of NaCl is +788 kJ `mol^(-1).` The enthalpies of hydration of `Na^(+)`(g) and `Cl^(-)` (g) ions are `-406 kJ mol^(-1)and-378 kJ mol^(-1)` respectively. The enthalpy of solution of NaCl(s) in water is

A

`786 kJ mol ^(-1)`

B

`4 kJ mol ^(-1)`

C

`- 4 kJ mol ^(-1)`

D

`- 792 kJ mol ^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(NaCl _((s)) to Na _((g) ) ^(+) + Cl _((g)) ^(-), Delta H =+ 788 kJ ("Lattice energy")),(Na _((g)) ^(+) H _(20 O _((l)) to Na _((aq)) ^(+)), Delta H =- 406 kJ ("Hydration")), (Cl _((g)) ^(-) + H _(2) O _((l)) to Cl _((aq)) ^(-) , Delta H =- 378kJ("Hydration")):}`
The net reaction of dissocultion of `Na Cl (s)` is `Na Cl _((s)) + 2 H _(2) O _((l)) to Na _((aq)) ^(+) + Cl _((aq)) ^(-)`
`Delta _(sol) H = 788 - ( 406 + 378) =- 4 kJ mol ^(-1)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CHEMICAL THERMODYNAMICS

    BRILLIANT PUBLICATION|Exercise LEVEL-II (Assertion-Reason Type)|18 Videos

  • CHEMICAL THERMODYNAMICS

    BRILLIANT PUBLICATION|Exercise LEVEL-III|52 Videos

  • CHEMICAL THERMODYNAMICS

    BRILLIANT PUBLICATION|Exercise LEVEL-I|92 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    BRILLIANT PUBLICATION|Exercise QUESTION (LEVEL -ll)|42 Videos

  • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

    BRILLIANT PUBLICATION|Exercise QUESTIONS (LEVEL-III ( Linked Comprehension Type ))|12 Videos

Similar Questions

Explore conceptually related problems

The lattice energy of NaCl is +788 kJ mol^(-1). The enthalpies of hydration of Na^(2)(g) and CI^(-) (g) ions are 406 kJ mol ^(-1) and-378 kJ mol^(-1) respectively. The enthalpy of solution of NaCl(s) in water is

Delta H_("combustion") of C _(6) H _(6) (l) . C(s) and H _(2) (g) are - 3000 kJ ,-390KJ mol^(-1) and - 290 kJ mol ^(-1) respectively. Hence heat of formation of benzene, is

Knowledge Check

  • The lattice energy of NaCl is 788 kJ mol^(-1) . This means that 788 kJ of energy is required

    A
    to separate one mole of solid NaCl into one mole of Na (9) and one mole of CI (g) to infinite distance
    B
    to separate one mole of solid NaCl into one mole of `Na^(+) (g)` and one mole of` Cl^(-)` (g) to infinite distance
    C
    to convert one mole of solid NaCl into one mole of gaseous Naci
    D
    to convert one mole of gaseous NaCl into one mole of solid NaCl

  • Ionisation energy of F^Ɵ is 320 kJ mol^(-1) . The electron gain enthalpy of fluorine would be

    A
    `-320 kJ mol^(-1)`
    B
    `-160 kJ mol^(-1)`
    C
    `+320 kJ mol^(-1)`
    D
    `+160 kJ mol^(-1)`

  • The enthalpy of combustion of H_(2)(g) at 298 K to give H_(2)O(g) is -249 kJ mol^(-1) and bond enthalpies of H-H and O=O are 433 kJ mol^(-1)and 492 kJ mol^(-1) respectively. The bond enthalpy ofO-His

    A
    `464 kJ mol ^(-1)`
    B
    `-464 kJ mol ^(-1)`
    C
    `232 kJ mol ^(-1)`
    D
    `-232 kJ mol ^(-1)`

    • Similar Questions

    Explore conceptually related problems

    The enthalpy of combustion of H_(2)(g)at 298 K to give H_(2) O (g) is - 279 kJ mol ^(-1) and bond enthalpies of H-H and O=O are 433 kJ mol^(-1) and 492 kJ mol^(-1) respectively. The bond enthalpy of O-H is

    The standard enthalpy of formation of NH_(3) is 46 kJ mol^(-1). If the enthalpy of formation of H_(2) from its atc is 436 kJ mol^(-1) and that of N_(2) is -712 kJ mol^(-1) the average bond enthalpy of N-H bond in NH_(3) is:

    a) The enthalpy of combustion of CH_(4(g)), C_(graphite) and H_(2(g)) at 298 K are - 890.3 kJ mol^-1 , - 393.5 kJ mol^-1 and -285.8 kJ mol^-1 respectively. Calculate the enthalpy of formation of CH_(4(g)).

    The standard enthalpy of formation of NH_(2) is -46 kJ mol^(-1). If the enthalpy of formation of H, from its atoms is -436 kJ mol^(-1) and that of N_(2) is - 712 kJ mol^(-1). the average bond enthalpy of N-H bond in NH_(3) is:

    Ionisation energy of F^Theta is 320 kJ "mol"^(-1) . The electron gain enthalpy of fluorine would be

    Home
    Profile