Question

The first and second ionisation potential of an element A are 246 and 439 kcal mol¹ respectively. What is the energy required for the following reaction?
A➡A^+2 + 2e-

​

Answer 1

Answer:

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Explanation:

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First and second ionization energies of Mg are 7.646 and 15.035 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into Mg

2+

ions present in 12 mg of magnesium vapours is:

[Given: 1 eV=96.5 kJ mol

−1

]

Hard

Solution

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Correct option is C)

The energy required to convert one Mg atom to Mg

2+

ion is the sum of first and second ionization potentials = 7.646+15.035=22.681eV

=96.5×22.681=2188.7kJ/mol

The atomic mass of Mg is 24 g/mol.

12 mg of Mg corresponds to 0.5 mmoles.

The amount of energy required to convert 0.5 mmol of Mg atoms to Mg

2+

ions = 0.5×10

−3

×2188.7=1.1kJ

Answer 2

Answer:

The energy required for the given reaction to occur is $685kcal/mol$.

Explanation:

Given,

The first ionisation potential of an element A, $I.E_{1}$ = $246kcal/mol$

The second ionisation potential of an element A, $I.E_{2}$ = $439kcal/mol$

The reaction has been given:

• $A \rightarrow A^{+2} + 2e^{-}$

The energy required for the given reaction to occur =?

The reaction process occurs in the way:

1)

• $A \rightarrow A^{+1} + 1e^{-}$       $I.E_{1} = 246kcal/mol$

2

• $A^{+1} \rightarrow A^{+2} + 1e^{-}$    $I.E_{2} = 439kcal/mol$

After adding both the equation, we get:

• $A \rightarrow A^{+2} + 2e^{-}$      $I.E = I.E_{1} + I.E_{2}$ = $246 + 439$ = $685kcal/mol$

Therefore, the energy required for the given reaction = $685kcal/mol$.