Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in $\text{kJ}/\text{mol}$.

Answer to Problem 104AP

Solution: $7.28\times {10}^3\text{kJ}/\text{mol}$

Explanation of Solution

Given information: $I_1=\text{520 kJ}/\text{mol}$

$E_n=-(2.18\times {10}^{-18}J)Z^2\left(\frac{1}{n^2}\right)$

$E=1.96\times {10}^4\text{kJ}/\text{mol}$

The total energy required to remove all the three electrons from lithium is given as: $E=I_1+I_2+I_3$ …… (1)

For lithium, the value of effective nuclear charge $\left(Z\right)$ is 3 and the value of principle quantum number $\left(n\right)$ is 1 because the third electron will come from the orbital of $1s$. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

$\begin{array}{c}{\mathit{\text{I}}}_3=\Delta \mathit{\text{E}}\\ ={\mathit{\text{E}}}_{\infty }-{\mathit{\text{E}}}_1\end{array}$

Or

$\begin{array}{c}{\mathit{\text{I}}}_3={\mathit{\text{E}}}_{\infty }-{\mathit{\text{E}}}_1\\ =-(2.18\times {10}^{-18}J)Z^2\left(\frac{1}{{\infty }^2}\right)--(2.18\times {10}^{-18}J)Z^2\left(\frac{1}{1^2}\right)\end{array}$

Substitute 3 for $Z$ in the above equation as follows:

$\begin{array}{c}{\mathit{\text{I}}}_3=-(2.18\times {10}^{-18}J){\left(3\right)}^2\left(\frac{1}{{\infty }^2}\right)-[-(2.18\times {10}^{-18}J){\left(3\right)}^2\left(\frac{1}{1^2}\right)]\\ =+1.96\times {10}^{-}{}^{17}J\end{array}$

In one mole, there are $6.022\times {10}^{23}$ atoms present.

So, the third ionization energy per mole is given as follows:

$\begin{array}{c}{I}_3=(1.96\times {10}^{-17}J)\times \left(\frac{6.022\times {10}^{23}}{1mol}\right)\\ =1.18\times {10}^7\text{J}/mol\end{array}$

Since $1\text{J}={\text{10}}^{-3}kJ$,

Therefore,

$\begin{array}{c}1.18\times {10}^4\text{J}/\text{mol}=\left(1.18\times {10}^4\frac{\cancel{\text{J}}}{\text{mol}}\right)\left({\text{10}}^{-3}\frac{kJ}{\cancel{\text{J}}}\right)\\ =1.18\times {10}^4\text{kJ}/\text{mol}\end{array}$

Now, substitute $1.18\times {10}^4\text{kJ}/\text{mol}$ for $I_3$, $520kJ/mol$ for $I_1$ and $1.96\times {10}^4\text{kJ}/\text{mol}$ for $E$ in equation (1) as follows:

$\begin{array}{c}1.96\times {10}^4\text{kJ}/\text{mol}=\left(520\text{ kJ}/\text{mol}\right)+I_2+\left(1.18\times {10}^4\text{kJ}/\text{mol}\right)\\ I_2=7.28\times {10}^3\text{kJ}/\text{mol}\end{array}$

Conclusion

The second ionization energy of lithium is $7.28\times {10}^3\text{kJ}/\text{mol}$.