# If lithium has a first ionisation energy of 520 kJ mol^(-1), estimate the energy required to form 1 Li^+ ion from 1 Li atom?

Nov 15, 2017

Around $8.64 \cdot {10}^{- 19} J = 5.40 e V$

#### Explanation:

The ionization energy tells us that for 1 mole of gaseous lithium, $520 k J$ of energy is needed to ionise each atom.

We also know that 1 mole contains $6.02 \cdot {10}^{23}$ atoms (to $3$ s.f.)

So, 1 atom = $\frac{1}{6.02 \cdot {10}^{23}} = 6.02 \cdot {10}^{- 23} m o l$

$520 k J = 520000 J = 5.2 \cdot {10}^{5} J$

$\left(5.2 \cdot {10}^{5}\right) \left(6.02 \cdot {10}^{- 23}\right) = 8.64 \cdot {10}^{- 19} J$

$\frac{8.64 \cdot {10}^{- 19}}{1.60 \cdot {10}^{- 19}} = 5.40 e V$

Nov 15, 2017

We are given the first ionization energy of the lithium atom....

#### Explanation:

That is we are given data for the reaction:

$L i \left(g\right) + \Delta \rightarrow L {i}^{+} \left(g\right) + {e}^{-}$...

Where $\Delta = 520 \cdot k J \cdot m o {l}^{-} 1$, and when we write $k J \cdot m o {l}^{-} 1$ we mean PER MOLE of reaction as written.....

And given the equation we need $520 \cdot k J$ to produce a mole of gaseous lithium cations....

To produce a SINGLE LITHIUM CATION, we divide the molar quantity by the Avocado number....the number of avocadoes in a mole, ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

$\frac{520 \cdot k J \cdot m o {l}^{-} 1}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1} = 8.64 \times {10}^{-} 22 \cdot k J = 8.64 \times {10}^{-} 19 \cdot J$. We could convert this value to $\text{ergs}$ or $c {m}^{-} 1$ or $e V$ if we are real masochists.