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This question is about lattice enthalpies - AQA - A-Level Chemistry - Question 1 - 2018 - Paper 1

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This question is about lattice enthalpies. Figure 1 shows a Born–Haber cycle for the formation of magnesium oxide. Complete Figure 1 by writing the missing symbols... show full transcript

Worked Solution & Example Answer:This question is about lattice enthalpies - AQA - A-Level Chemistry - Question 1 - 2018 - Paper 1

Step 1

Complete the Born–Haber Cycle with the Missing Symbols

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Answer

  1. For the ionization of magnesium (Mg): Write the reaction as follows:
ightarrow ext{Mg}^{2+}(g) + 2e^-$$ 2. For the dissociation of oxygen (O): The reaction will be: $$ rac{1}{2} ext{O}_2(g) ightarrow ext{O}(g) + e^-$$ 3. For the formation of magnesium oxide (MgO): Place it at the bottom as: $$ ext{Mg}(s) + ext{O}(g) ightarrow ext{MgO}(s)$$

Step 2

Calculate the Enthalpy Change for the Formation of Magnesium Oxide

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Answer

We will use Hess's law summarizing the enthalpy changes from the given data in Table 1:

ΔHf(MgO)=ΔHat(Mg)+IE1(Mg)+IE2(Mg)+BE1(O2)+EA1(O)+EA2(O)+ΔHf(MgO)\Delta H_f\text{(MgO)} = \Delta H_{at} \text{(Mg)} + \text{IE}_1 \text{(Mg)} + \text{IE}_2 \text{(Mg)} + \text{BE}_{1}(O_2) + \text{EA}_1\text{(O)} + \text{EA}_2\text{(O)} + \Delta H_f \text{(MgO)}

Substituting values:

ΔHf(MgO)=150+736+1450+495+(142)+(484)+602\Delta H_f\text{(MgO)} = 150 + 736 + 1450 + 495 + (-142) + (-484) + 602

Calculating gives:

ΔHf=150+736+1450+495142484+602=3888/380=10.21 kJ mol1\Delta H_f = 150 + 736 + 1450 + 495 - 142 - 484 + 602 = -3888/ 380 = -10.21 \text{ kJ mol}^{-1}.

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