Use IE and EA listed below to determine whether the following process is endothermic or exothermic- Mg g - 2F g - Mg2- g - 2F- g IE1 of Mgg - 737-7 kJ mol-1 IE2 of Mgg - 1451 kJ mol-1 EA of Fg -328 kJ mol-1

The correct option is A Endothermic nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp ...

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Byju's Answer

Standard XII

Chemistry

Trend in Ionisation Enthalpy

Use IE and EA...

Question

Use (IE) and (EA) listed below to determine whether the following process is endothermic or exothermic:
$M g (g) + 2 F (g) \to M g^{2 +} (g) + 2 F^{-} (g)$
$(I E)_{1}$ of $M g (g) = 737.7 k J m o l^{- 1}$
$(I E)_{2}$ of $M g (g) = 1451 k J m o l^{- 1}$
$(E A)$ of $F (g) = - 328 k J m o l^{- 1}$

Endothermic

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Exothermic

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Isothermic

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None of these

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Solution

The correct option is A Endothermic
$Δ H$
$M g (g) \to M g^{+} (g) + e^{-}$ 737.7 kJ
$M g^{+} (g) \to M g^{2 +} (g) + e^{-}$ 1451 kJ
$2 F (g) + 2 e^{-} \to 2 F^{-} (g)$ -328 $\times$ 2kJ
$M g (g) + 2 F (g) \to M g^{2 +} (g) + 2 F^{-} (g) Δ H = 1532.7 k J$
Thus, endothermic

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Similar questions

Q. Using the data given below, predict

Δ H

for the reaction in

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

M g (g) + 2 F (g) ⟶ {M g}^{2 +} (g) + 2 F^{-} (g)

{I E}_{2}

and

{I E}_{2}

M g (g)

are

737.7

and

1451 k J

{m o l}^{- 1}

{E A}_{1}

for

F (g)

- 328 k J

{m o l}^{- 1}

[Take the nearest integer]

I E_{1}

and

I E_{2}

of magnesium are

178

and

348

kcal/mole.

The energy for the reaction

M g_{(g)} \to M g_{(g)}^{2 +} + 2 e^{-}

is:

Q. Calculate the enthalpy of formation of

M g F_{2}

from the following data-
Enthalpy of sublimation for

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

Enthalpy of dissociation of

F_{2} (g) \to 2 F (g) = 155.8 k J m o l^{- 1}

Ionisation energy of

M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

Electron gain enthalpy of

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

Lattice enthalpy of

M g F_{2} = - 2922.5 k J m o l^{- 1}

Q. Calculate the enthalpy of formation of

M g F_{2}

from thefollowing data-
Enthalpy of sublimation for

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

Enthalpy of dissociation of

F_{2} (g) \to 2 F (g) = 155.8 k J m o l^{- 1}

Ionisation energy of

M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

Electron gain enthalpy of

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

Lattice enthalpy of

M g F_{2} = - 2922.5 k J m o l^{- 1}

Q. The

△_{i} H_{1}

and

△_{i} H_{2}

of Mg(g) are 740 and 1450 kJ

△_{i} H_{2}

{m o l}^{- 1}

. Calculate the percentage of

{M g}^{+} (g)

and

{M g}^{2 +} (g)

if 1 g of Mg(g) absorbs 50 kJ of energy.

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Use (IE) and (EA) listed below to determine whether the following process is endothermic or exothermic:Mg(g)+2F(g)→Mg2+(g)+2F−(g)(IE)1 of Mg(g)=737.7kJmol−1(IE)2 of Mg(g)=1451kJmol−1(EA) of F(g)=−328kJmol−1

The correct option is A Endothermic ΔHMg(g)→Mg+(g)+e− 737.7 kJMg+(g)→Mg2+(g)+e− 1451 kJ2F(g)+2e−→2F−(g) -328 × 2kJMg(g)+2F(g)→Mg2+(g)+2F−(g)ΔH=1532.7kJThus, endothermic

Use (IE) and (EA) listed below to determine whether the following process is endothermic or exothermic:
$M g (g) + 2 F (g) \to M g^{2 +} (g) + 2 F^{-} (g)$
$(I E)_{1}$ of $M g (g) = 737.7 k J m o l^{- 1}$
$(I E)_{2}$ of $M g (g) = 1451 k J m o l^{- 1}$
$(E A)$ of $F (g) = - 328 k J m o l^{- 1}$

The correct option is A Endothermic
$Δ H$
$M g (g) \to M g^{+} (g) + e^{-}$ 737.7 kJ
$M g^{+} (g) \to M g^{2 +} (g) + e^{-}$ 1451 kJ
$2 F (g) + 2 e^{-} \to 2 F^{-} (g)$ -328 $\times$ 2kJ
$M g (g) + 2 F (g) \to M g^{2 +} (g) + 2 F^{-} (g) Δ H = 1532.7 k J$
Thus, endothermic