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The energy needed for {L}{i}_{{{g}}}{L}{{i}_{{{g}}}^{{{3}+}}}+

Question

The energy needed for $L i_{g} \to L i_{g}^{3 +} + 3 e$ is $1.96 \times 10^{4} k J m o l^{- 1}$ . If the first ionisation energy of Li is $520 k J m o l^{- 1}$ . Calcuate the second ionisation energy of $L i$ .
(Given : $I E_{1}$ for $H = 2.2.18 \times 10^{- 18} k J atom^{- 1}$ ).

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The energy needed for

L i_{g} \to L i_{g}^{3 +} + 3 e

1.96 \times 10^{4} k J m o l^{- 1}

. If the first ionisation energy of Li is

520 k J m o l^{- 1}

. Calcuate the second ionisation energy of

L i

.
(Given :

I E_{1}

for

H = 2.2.18 \times 10^{- 18} k J atom^{- 1}

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Text solutionVerified

Solution: Given

L i (g) \to L i_{g}^{3 +} + 3 d, Δ H = 1.96 \times 10^{4} k J mol^{- 1}

….(1)

L i_{g} \to L i_{g}^{+} + e, I E_{1} = 520 k J mole^{- 1}

….(2)

L i_{g} \to L i_{g}^{+} + e, I E_{2} = a k J mol^{- 1}

…..(3)

L i_{g} \to L i_{g}^{3 +} + e, I E_{3} = b k J mole^{- 1} \dots (4)

Also

b = E_{1}

for

L i^{2 +} = E_{1}

for

H \times Z^{2}

= 2.18 \times 10^{- 18} \times 3^{2} \times 6.023 \times 10^{23} J mole^{- 1}

= 11.81 X X 10^{6} j M O L^{1}

= 11.881 \times 10^{3} k J mol^{- 1}

By eqs , (1) , (2) , (3) and (4),

1.96 \times 10^{4} = 520 + a + 11.9 = 81 \times 10^{3}

a = 7270 k J mol^{- 1}

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Question Text	The energy needed for $L i_{g} \to L i_{g}^{3 +} + 3 e$ is $1.96 \times 10^{4} k J m o l^{- 1}$ . If the first ionisation energy of Li is $520 k J m o l^{- 1}$ . Calcuate the second ionisation energy of $L i$ . (Given : $I E_{1}$ for $H = 2.2.18 \times 10^{- 18} k J atom^{- 1}$ ).
Answer Type	Text solution:1
Upvotes	150