The ionization enthalpy of Na ^ + formation from Na _ (g) is 495 . 8kJ mol ^ - 1 , while the electron gain enthalpy of Br is - 325 . 0kJ mol ^ - 1 . Given the lattice enthalpy of NaBr is - 728 . 4kJ mol ^ - 1 . The

Correct answer: 5576 Δ H _ formation = IE _ 1 + Δ Heg _ 1 + LE = 495 . 8 + (- 325) + (- 728 . 4) = - 557 . 6 = - 5576 × 10 ^ - 1 KJ / mol . Note: The above calculation is not for Δ H _ formation but for Δ H _ Reaction But on the basis of given data it is the best ans.

JEE Main 202125 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2021 · 25 February · Morning Shift · Chemistry · Thermodynamics (C)

The ionization enthalpy of Na ^ + formation from Na _ (g) is 495 ....

The ionization enthalpy of

{Na}^{+}

formation from

{Na}_{\left(g\right)}

495 . 8kJ{mol}^{- 1},

while the electron gain enthalpy of

Br

- 325 . 0kJ{mol}^{- 1}

. Given the lattice enthalpy of

NaBr

- 728 . 4kJ{mol}^{- 1}

. The energy for the formation of

NaBr

ionic solid is \left(-\right)_____{10}^{- 1}kJ{mol}^{- 1} The ionization enthalpy of Na ^ + formation from Na _ (g) is 495 . 8kJ mol ^ - 1 , while the electron gain enthalpy of

Correct answer

5576

Step-by-step solution

The ionization enthalpy of Na ^ + formation from Na _ (g) is 495 . 8kJ mol ^ - 1 , while the electron gain enthalpy of

${\Delta H}_{formation} = {IE}_{1} + {\Delta Heg}_{1} + LE$

$= 495 . 8 + \left(- 325\right) + \left(- 728 . 4\right)$

$= - 557 . 6$

$= - 5576 \times {10}^{- 1}KJ / mol .$

Note: The above calculation is not for

${\Delta H}_{formation}$ but for ${\Delta H}_{Reaction}$

But on the basis of given data it is the best ans.

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