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Concept explainers
(a)
Interpretation:
ΔH°f of MgCl is to be calculated.
Concept introduction:
Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn
Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.
(a)
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Answer to Problem 9.75P
ΔH°f of MgCl is −125 kJ.
Explanation of Solution
The enthalpy change of the following reaction is ΔH°1.
Mg(s)→Mg(g) (1)
The enthalpy change of the following reaction is ΔH°2.
Cl2(g)→2Cl(g) (2)
The enthalpy change of the following reaction is ΔH°3.
Mg(g)→Mg+(g)+e− (3)
The enthalpy change of the following reaction is ΔH°4.
Cl(g)+e−→Cl−(g) (4)
The enthalpy change of the following reaction is ΔH°lattice.
MgCl(s)→Mg+(g)+Cl−(g) (5)
Reverse the equation (5).
Mg+(g)+Cl−(g)→MgCl(s) (6)
The enthalpy change for the reaction (6) is calculated as,
ΔH°5=−ΔH°lattice
Multiply the equation (2) by 12.
12Cl2(g)→Cl(g) (7)
The enthalpy change for the reaction (7) is calculated as,
ΔH°6=12ΔH°2
Add equation (1),(3), (4), (6) and (7).
Mg(s)+12Cl2(g)→MgCl(s) (8)
The enthalpy change of the final reaction (8) is ΔH°f.
The expression to calculate ΔH°f is as follows:
ΔH°f=ΔH°1+ΔH°3+ΔH°4+ΔH°5+ΔH°6 (9)
Substitute 148 kJ for ΔH°1 and 121.5 kJ for ΔH°3, 768 kJ for ΔH°4, −349 kJ for ΔH°5 and −783.5 kJ for ΔH°6 n the equation (9).
ΔH°f=148 kJ+121.5 kJ+768 kJ+(−349 kJ)+(−783.5 kJ)=−125 kJ
ΔH°f of MgCl is −125 kJ.
(b)
Interpretation:
Whether MgCl is favoured energetically relative to Mg and Cl2 is to be determined.
Concept introduction:
The standard enthalpy of formation (ΔH°f) is the enthalpy change of the formation of 1 mol of compound from its elements and the reactants and products, in this case, is also present in their standard state.
Standard state includes 1 atm pressure for gases, 1 M concentration for substances in the aqueous phase and the most stable form of substance at 1 atm pressure and 298 K temperature. For example, metal like sodium exists in solid state and molecular elements like halogen exist in a gaseous state.
(b)
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Answer to Problem 9.75P
MgCl is favoured energetically relative to Mg and Cl2 due to the negative value of heat of formation.
Explanation of Solution
The chemical equation for the formation of MgCl is as follows:
Mg(s)+12Cl2(g)→MgCl(s)
The enthalpy of formation (ΔH°f) of MgCl is −125 kJ.
When the enthalpy of formation is negative then the compound formed is relatively more stable as compared to the elements. Therefore, MgCl is relatively more stable as compared to Mg and Cl2.
When the enthalpy of formation is negative then the compound formed is relatively more stable as compared to the elements.
(c)
Interpretation:
ΔH° for the conversion of MgCl to MgCl2 and Mg is to be calculated.
Concept introduction:
The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔH°rxn) is as follows:
ΔH°rxn=∑mΔH°f (products)−∑mΔH°f (reactants)
Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.
(c)
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Answer to Problem 9.75P
ΔH° for the conversion of MgCl to MgCl2 and Mg is −392 kJ.
Explanation of Solution
The chemical equation for the formation of MgCl2 is as follows:
2MgCl(s)→MgCl2(s)+Mg(s)
The formula to calculate the standard enthalpy of reaction (ΔH°) is as follows:
ΔH°=[{ΔH°f[MgCl2(s)]+ΔH°f[Mg(s)]}−{2ΔH°f[MgCl(s)]}] (10)
Substitute −641.6 kJ/mol for ΔH°f[MgCl2(s)], 0 for ΔH°f[Mg(s)] and −125 kJ for ΔH°f[MgCl(s)] in the equation (10).
ΔH°=[{(1 mol)(−641.6 kJ/mol)+(1 mol)(0)}−{(2 mol)(−125 kJ)}]=−391.6 kJ≈−392 kJ
ΔH° for the conversion of MgCl to MgCl2 and Mg is −392 kJ.
(d)
Interpretation:
Whether MgCl is favoured energetically relative to MgCl2 is to be determined.
Concept introduction:
The standard enthalpy of formation (ΔH°f) is the enthalpy change of the formation of 1 mol of compound from its elements and the reactants and products, in this case, is also present in their standard state.
Standard state includes 1 atm pressure for gases, 1 M concentration for substances in the aqueous phase and the most stable form of substance at 1 atm pressure and 298 K temperature. For example, metal like sodium exists in solid state and molecular elements like halogen exist in a gaseous state.
(d)
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Answer to Problem 9.75P
MgCl2 is favoured energetically relative to MgCl due to the more negative value of heat of formation of MgCl2 as compared to MgCl.
Explanation of Solution
The chemical equation for the formation of MgCl2 is as follows:
2MgCl(s)→MgCl2(s)+Mg(s)
The enthalpy of formation (ΔH°f) of MgCl is −125 kJ.
The enthalpy of formation (ΔH°f) of MgCl2 is −392 kJ.
The compound with the more negative enthalpy of formation is relatively more stable than the compound with the less negative enthalpy of formation. Therefore, MgCl2 is relatively more stable as compared to MgCl.
The compound with the more negative enthalpy of formation is relatively more stable as compared to the compound with the less negative enthalpy of formation.
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