# Ionization energy

1. Atomic potassium emits light very weakly at 365 nm when an excited electron moves from a 4d-orbital to a 4s-orbital , and at 689 nm when an electron moves from a 4d-orbital to a 4p-orbital. What is the energy separation (in kJ/mol) between the 4s- and 4p-orbitals in a potassium atom?

2. Look up the ionization energy for hydrogen, in kJ/mol (cite your source; this value is widely available, but can also be found in chapter 8 of your textbook). Compare this quantity with the value determined by calculating the ΔE for an electron transition from n = 1 to n = ∞ using the Bohr formula for the energy levels (express the result in kJ/mol).

Ionization Energy for H: ________ (Source:______________) Calculated value:_____

3. A student mixed 50Ml of water containing 0.010moles of HCl with 60Ml of water containing 0.012 moles of NaOH in a coffee cup calorimeter. Both the solutions were initially 25C but after reaction upon mxing the temperature rose to 26.2C. The specific heat of the solution is 4.18J/g C, and the calorimeter has a het capacity of 35J/C. The reaction is shown below. How much heat is produced in the reaction? How much heat is produced if one mole of water is formed in this chemical reaction?

1. the separation energy between levels 4d and 4s is

$E(d-s) = h*F_1 = (h*c)/\lambda_1 = (6.626*10^{-34})*(3*10^8)/(365*10^{-9}) =5.44*10^{-19} J$

the separation energy between levels 4d and 4p is

$E(d-p) = h*c/\lambda_2 = (6.626*10^{-34})*(3*10^8)/(689*10^{-9}) =2.885*10^{-19} J$

The separation energy between 4p and 4s is

$E(p-s) =5.44*10^{-19} -2.885*10^{-19} =2.55*10^{-19} J =1.597 eV$

For Hydrogen $E=1312 KJ/mol$

in one mol there are Avogadro number atoms. $N_A = 6.023*10^{23} atoms$

Energy per atom $E =1312*10^3/6.023*10^{23} =2.178*10^{-18} (J/atom)$

From Bohr model of atom transition energy between states n and m is

$E(n-m) = E_1 *(1/n^2 – 1/m^2)$ where $E_1=-13.6 eV$ is the energy of first level

Therefore the ionization energy for Hydrogen is ($n= \infty$ , $m=1$) from Bohr model is

$E = 13.6 eV = 13.6*1.6*10^{-19} =2.176*10^{-18} J$

For one mole the energy is

$E(one mole) = E*N_A =2.176*10^{-18}*6.023*10^{23} =1310.6 KJ/mol$

Therefore

Ionization Energy for H: 1312 KJ/mol

Calculated value: 1310.6 KJ/mol

3.

$m(solution)*c(solution)(Tf-Ti) + C(calorimeter)*(Tf-Ti) = Q$

We suppose the solution has density of 1 g/ml

$Q = (50+60)*4.18*(26.2-25) +35*(26.2-25) =593.76 J$

The eq. of reaction between HCl and NaOH is

$HCl +NaOH = H2O + NaC$

from here one can see that to form one mole of water it is necessary one mole of HCl and one mole of NaOH . Therefore in the reaction need to enter not 0.010 mol HCl and 0.012 mol NaOH but 1 mol HCl and 1 mol NaOH.

Heat of reaction is

$H = H(products) – H(reactants)$

$H(water) = 285.8 KJ/mol$

$H(NaCl) =385.92 KJ/mol$

$H(HCl) = 92.30 kJ/mol$

$H(NaOH) =469.15 KJ/mol$

$H = (285.8+385.92)-(469.15+92.30) =110.27 KJ/mol$

Since it is formed one mole of water the total heat produced is approximately equal to

$Q =H = 110.27 KJ$