Examkin

Soru

Zorluk: Çok zorStandard Enthalpy Changes and Hess's Law
Given the following thermochemical equations:
I. 2Fe(s)+32O2(g)→Fe2O3(s)ΔH∘=−824.2 kJ mol−1\text{I. } 2\text{Fe}(s) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(s) \quad \Delta H^\circ = -824.2\text{ kJ mol}^{-1}
II. CO(g)+12O2(g)→CO2(g)ΔH∘=−283.0 kJ mol−1\text{II. } \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -283.0\text{ kJ mol}^{-1}
What is the standard enthalpy change, ΔH∘\Delta H^\circ, for the reduction of iron(III) oxide by carbon monoxide according to the following reaction?
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)
  1. −24.8 kJ mol−1-24.8\text{ kJ mol}^{-1}Cevap
  2. B
    −1673.2 kJ mol−1-1673.2\text{ kJ mol}^{-1}
  3. C
    +1673.2 kJ mol−1+1673.2\text{ kJ mol}^{-1}
  4. D
    +541.2 kJ mol−1+541.2\text{ kJ mol}^{-1}

Cevap

−24.8 kJ mol−1-24.8\text{ kJ mol}^{-1}
According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for individual intermediate steps. Reversing the formation equation of Fe2O3(s)\text{Fe}_2\text{O}_3(s) flips its enthalpy sign from −824.2 kJ mol−1-824.2\text{ kJ mol}^{-1} to +824.2 kJ mol−1+824.2\text{ kJ mol}^{-1}. Multiplying the oxidation reaction of CO(g)\text{CO}(g) by 3 scales its enthalpy from −283.0 kJ mol−1-283.0\text{ kJ mol}^{-1} to −849.0 kJ mol−1-849.0\text{ kJ mol}^{-1}. Adding both modified values yields +824.2 kJ mol−1−849.0 kJ mol−1=−24.8 kJ mol−1+824.2\text{ kJ mol}^{-1} - 849.0\text{ kJ mol}^{-1} = -24.8\text{ kJ mol}^{-1}.

Adım Adım Çözüm

1
Reverse Equation I so that Fe2O3(s)\text{Fe}_2\text{O}_3(s) becomes a reactant.
Fe2O3(s)→2Fe(s)+32O2(g)ΔH1∘=+824.2 kJ mol−1\text{Fe}_2\text{O}_3(s) \rightarrow 2\text{Fe}(s) + \frac{3}{2}\text{O}_2(g) \quad \Delta H_1^\circ = +824.2\text{ kJ mol}^{-1}
Reversing a thermochemical equation changes the sign of its standard enthalpy change, ΔH∘\Delta H^\circ.
2
Multiply Equation II by 3 so that the stoichiometric coefficient of CO(g)\text{CO}(g) matches the target reaction.
3CO(g)+32O2(g)→3CO2(g)ΔH2∘=3×(−283.0 kJ mol−1)=−849.0 kJ mol−13\text{CO}(g) + \frac{3}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) \quad \Delta H_2^\circ = 3 \times (-283.0\text{ kJ mol}^{-1}) = -849.0\text{ kJ mol}^{-1}
Enthalpy change is an extensive property, so multiplying reaction coefficients requires multiplying ΔH∘\Delta H^\circ by the same factor.
3
Sum the modified thermochemical equations according to Hess's Law.
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)
ΔH∘=ΔH1∘+ΔH2∘=+824.2 kJ mol−1+(−849.0 kJ mol−1)=−24.8 kJ mol−1\Delta H^\circ = \Delta H_1^\circ + \Delta H_2^\circ = +824.2\text{ kJ mol}^{-1} + (-849.0\text{ kJ mol}^{-1}) = -24.8\text{ kJ mol}^{-1}
Hess's Law states that the overall enthalpy change for a chemical reaction is independent of the pathway taken.

Anahtar Kavram

Hess's Law of Constant Heat Summation
Tahmini Süre:2m 0s
Bu soruyu puanla

Konular

Standard Enthalpy Changes and Hess's Law