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Zorluk: OrtaStandard Enthalpy Changes and Hess's Law
Consider the reduction of zinc oxide by carbon monoxide:
ZnO(s)+CO(g)→Zn(s)+CO2(g)\text{ZnO}(s) + \text{CO}(g) \rightarrow \text{Zn}(s) + \text{CO}_2(g)
Given the standard enthalpies of formation (ΔHf∘\Delta H_f^\circ):
- ΔHf∘[ZnO(s)]=−348.0 kJ mol−1\Delta H_f^\circ[\text{ZnO}(s)] = -348.0\text{ kJ mol}^{-1}
- ΔHf∘[CO(g)]=−110.5 kJ mol−1\Delta H_f^\circ[\text{CO}(g)] = -110.5\text{ kJ mol}^{-1}
- ΔHf∘[CO2(g)]=−393.5 kJ mol−1\Delta H_f^\circ[\text{CO}_2(g)] = -393.5\text{ kJ mol}^{-1}

What is the standard enthalpy change of the reaction, ΔH∘\Delta H^\circ, in kJ mol−1\text{kJ mol}^{-1}?

Cevap: 65 kJ mol^-1

Cevap

The standard enthalpy change of the reaction is +65.0 kJ mol^-1.
According to Hess's Law, the standard enthalpy change of a reaction is calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. For this reaction, ΔH∘=[−393.5+0]−[−348.0+(−110.5)]=−393.5−(−458.5)=+65.0 kJ mol−1\Delta H^\circ = [-393.5 + 0] - [-348.0 + (-110.5)] = -393.5 - (-458.5) = +65.0\text{ kJ mol}^{-1}.

Adım Adım Çözüm

1
Determine the standard enthalpy of formation for zinc element in standard state
ΔHf∘[Zn(s)]=0 kJ mol−1\Delta H_f^\circ[\text{Zn}(s)] = 0\text{ kJ mol}^{-1}
By definition, the standard enthalpy of formation of an element in its standard reference state is zero.
2
Apply Hess's Law relationship using enthalpies of formation
ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})
Enthalpy change of a reaction equals the total enthalpy of formation of products minus that of reactants.
3
Calculate the sum of formation enthalpies for products
∑ΔHf∘(products)=−393.5+0=−393.5 kJ mol−1\sum \Delta H_f^\circ(\text{products}) = -393.5 + 0 = -393.5\text{ kJ mol}^{-1}
Products are 1 mole of CO2(g)\text{CO}_2(g) and 1 mole of Zn(s)\text{Zn}(s).
4
Calculate the sum of formation enthalpies for reactants
∑ΔHf∘(reactants)=−348.0+(−110.5)=−458.5 kJ mol−1\sum \Delta H_f^\circ(\text{reactants}) = -348.0 + (-110.5) = -458.5\text{ kJ mol}^{-1}
Reactants are 1 mole of ZnO(s)\text{ZnO}(s) and 1 mole of CO(g)\text{CO}(g).
5
Subtract reactant total from product total to obtain reaction enthalpy
ΔH∘=−393.5−(−458.5)=+65.0 kJ mol−1\Delta H^\circ = -393.5 - (-458.5) = +65.0\text{ kJ mol}^{-1}
Performing the subtraction −393.5+458.5-393.5 + 458.5 yields +65.0 kJ mol−1+65.0\text{ kJ mol}^{-1}.

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Calculating standard enthalpy change of reaction using standard enthalpies of formation via Hess's Law.
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Standard Enthalpy Changes and Hess's Law