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Zorluk: ZorStandard Enthalpy Changes and Hess's Law
The hydration of ethene to produce liquid ethanol is represented by the chemical equation:
C2H4(g)+H2O(g)→C2H5OH(l)\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)

Given the following thermochemical equations:
1. C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)ΔH∘=−1367 kJ mol−1\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \quad \Delta H^\circ = -1367\text{ kJ mol}^{-1}
2. C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)ΔH∘=−1411 kJ mol−1\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H^\circ = -1411\text{ kJ mol}^{-1}
3. H2O(g)→H2O(l)ΔH∘=−44 kJ mol−1\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -44\text{ kJ mol}^{-1}

Calculate the standard enthalpy change, ΔH∘\Delta H^\circ, for the hydration reaction in kJ mol−1\text{kJ mol}^{-1}.

Cevap: -88 kJ/mol

Cevap

The standard enthalpy change for the hydration reaction is -88 kJ/mol.
According to Hess's law, the standard enthalpy change of a net reaction can be determined by algebraically combining component reactions and their enthalpy values. Adding Equation 2 as written, Equation 3 as written, and the reverse of Equation 1 cancels out intermediate species (carbon dioxide, oxygen, and liquid water), leaving the net hydration reaction. Summing their respective enthalpy values gives -1411 kJ/mol + (-44 kJ/mol) + 1367 kJ/mol = -88 kJ/mol.

Adım Adım Çözüm

1
Identify the required target thermochemical equation
Target: C2H4(g)+H2O(g)→C2H5OH(l)\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)
This establishes the stoichiometry and physical states required for reactants and products.
2
Apply Hess's law to reverse and combine the given thermochemical equations
Keep Equation 2: ΔH2=−1411 kJ mol−1\Delta H_2 = -1411\text{ kJ mol}^{-1}
Keep Equation 3: ΔH3=−44 kJ mol−1\Delta H_3 = -44\text{ kJ mol}^{-1}
Reverse Equation 1: ΔH1′=+1367 kJ mol−1\Delta H_1' = +1367\text{ kJ mol}^{-1}
Reversing Equation 1 places liquid ethanol on the product side, requiring the sign of its enthalpy change to be inverted.
3
Sum the enthalpy changes of the modified reaction steps
ΔH∘=(−1411)+(−44)+(+1367)=−88 kJ mol−1\Delta H^\circ = (-1411) + (-44) + (+1367) = -88\text{ kJ mol}^{-1}
According to Hess's law, the total enthalpy change of an overall reaction equals the sum of the enthalpy changes for individual component steps.

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Standard Enthalpy Changes and Hess's Law