Question

question 10
0.5 pts
given the following data, determine the enthalpy of formation for mgo (s) in kj/mol.
δh_bond,o₂ (g) = +498 kj/mol
e_ea1 = first electron affinity for o (g) = -142 kj/mol
e_ea2 = second electron affinity for o (g) = +844 kj/mol
e_i1 = first ionization energy for mg (g) = +738.2 kj/mol
e_i2 = second ionization energy for mg (g) = +1449.8 kj/mol
δh_sublimation = heat of sublimation for mg (s) = +148.3 kj/mol
δh_lattice = lattice energy for mgo (s) = +3890.1 kj/mol
report the answer in kilojoules per mole to the correct sig figs, but do not include units in your answer.

Explanation:

Step1: Recall Born - Haber Cycle

The enthalpy of formation ($\Delta H_f$) of a compound can be determined using the Born - Haber cycle. The formula for the enthalpy of formation of $\ce{MgO(s)}$ is:
$\Delta H_f=\Delta H_{\text{sublimation}}+\ E_{I1}+E_{I2}+\frac{1}{2}\Delta H_{\text{bond},\ce{O2(g)}}+E_{\text{ea1}}+E_{\text{ea2}}-\Delta H_{\text{lattice}}$

Step2: Substitute the given values

• $\Delta H_{\text{sublimation}} = 148.3\ \text{kJ/mol}$
• $E_{I1}=738.2\ \text{kJ/mol}$
• $E_{I2}=1449.8\ \text{kJ/mol}$
• $\frac{1}{2}\Delta H_{\text{bond},\ce{O2(g)}}=\frac{1}{2}\times498 = 249\ \text{kJ/mol}$
• $E_{\text{ea1}}=- 142\ \text{kJ/mol}$
• $E_{\text{ea2}} = 844\ \text{kJ/mol}$
• $\Delta H_{\text{lattice}}=3890.1\ \text{kJ/mol}$

Now substitute these values into the formula:

$\Delta H_f=148.3 + 738.2+1449.8 + 249-142 + 844-3890.1$

First, add the positive terms: $148.3+738.2 = 886.5$; $886.5 + 1449.8=2336.3$; $2336.3+249 = 2585.3$; $2585.3+844=3429.3$

Then add the negative terms: $- 142-3890.1=-4032.1$

Now combine the results: $3429.3-4032.1=-602.8$

Answer:

-603 (rounded to the correct significant figures)