The first I-E- of Li is 5-41 eV and first electron gain enthalpy of Cl is -3-61 eV- Calculate - H in kJ mol -1 for the reaction-Li g - Cl g - Li g - Cl g -A- 347-4 kJB- 173-7 kJC- 133-6 kJD- 193-2 kJ

The correct option is B 173.7 kJExplanation : nbsp; Δ H/molecule=IE+EA=5.41 3.61= 1.80 eV Δ H/mole=1.8× 6.022×10^23eV =1.8×6.022×10^23×1.602×10^ 19×10^ 3 ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The first I.E...

Question

The first I.E. of $L i$ is 5.41 eV and first electron gain enthalpy of $C l$ is - 3.61 eV. Calculate $Δ H$ in $k J m o l^{- 1}$ for the reaction:
$L i_{(g)} + C l_{(g)} \to L i_{(g)}^{+} + C l_{(g)}^{-}$

133.6 kJ

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347.4 kJ

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173.7 kJ

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193.2 kJ

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Solution

The correct option is C 173.7 kJ
Explanation :

$\frac{Δ H}{molecule} = I E + E A = 5.41 - 3.61 = 1.80 e V$

$\frac{Δ H}{mole} = 1.8 \times 6.022 \times 10^{23} e V$

$= 1.8 \times 6.022 \times 10^{23} \times 1.602 \times 10^{- 19} \times 10^{- 3}$

$= 173.7 k J$

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The first I.E. of Li is 5.41 eV and first electron gain enthalpy of Cl is - 3.61 eV. Calculate ΔH in kJmol−1 for the reaction: Li(g)+Cl(g)→Li+(g)+Cl−(g)

The correct option is C 173.7 kJExplanation : ΔHmolecule=IE+EA=5.41−3.61=1.80eV ΔHmole=1.8×6.022×1023eV =1.8×6.022×1023×1.602×10−19×10−3 =173.7 kJ

The first I.E. of $L i$ is 5.41 eV and first electron gain enthalpy of $C l$ is - 3.61 eV. Calculate $Δ H$ in $k J m o l^{- 1}$ for the reaction:
$L i_{(g)} + C l_{(g)} \to L i_{(g)}^{+} + C l_{(g)}^{-}$

The correct option is C 173.7 kJ
Explanation :

$\frac{Δ H}{molecule} = I E + E A = 5.41 - 3.61 = 1.80 e V$

$\frac{Δ H}{mole} = 1.8 \times 6.022 \times 10^{23} e V$

$= 1.8 \times 6.022 \times 10^{23} \times 1.602 \times 10^{- 19} \times 10^{- 3}$

$= 173.7 k J$