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Zorluk: OrtaStandard Enthalpy Changes and Hess's Law
Consider the following standard enthalpies of combustion at 298 K298\text{ K}:
C(s)+O2(g)→CO2(g)ΔH∘=−394 kJ mol−1\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -394\text{ kJ mol}^{-1}
H2(g)+12O2(g)→H2O(l)ΔH∘=−286 kJ mol−1\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -286\text{ kJ mol}^{-1}
C2H2(g)+52O2(g)→2CO2(g)+H2O(l)ΔH∘=−1300 kJ mol−1\text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(l) \quad \Delta H^\circ = -1300\text{ kJ mol}^{-1}

Using Hess's law, calculate the standard enthalpy of formation of ethyne gas, C2H2(g)\text{C}_2\text{H}_2(g), in kJ mol−1\text{kJ mol}^{-1}.

Cevap: 226 kJ mol^{-1}

Cevap

The standard enthalpy of formation of ethyne gas is +226 kJ mol−1+226\text{ kJ mol}^{-1} (or 226 kJ mol−1226\text{ kJ mol}^{-1}).
Applying Hess's law involves expressing the enthalpy of formation of ethyne as the sum of twice the enthalpy of combustion of carbon, once the enthalpy of combustion of hydrogen, minus the enthalpy of combustion of ethyne: ΔHf∘=2(−394)+(−286)−(−1300)=−788−286+1300=+226 kJ mol−1\Delta H_f^\circ = 2(-394) + (-286) - (-1300) = -788 - 286 + 1300 = +226\text{ kJ mol}^{-1}.

Adım Adım Çözüm

1
Write the standard formation equation for ethyne
2C(s)+H2(g)→C2H2(g)2\text{C}(s) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_2(g)
The enthalpy of formation represents the formation of one mole of a compound from its constituent elements in their standard states.
2
Manipulate given thermochemical equations to match the target equation
Multiply equation 1 by 2 (ΔH=−788 kJ\Delta H = -788\text{ kJ}), keep equation 2 unchanged (ΔH=−286 kJ\Delta H = -286\text{ kJ}), and reverse equation 3 (ΔH=+1300 kJ\Delta H = +1300\text{ kJ})
According to Hess's Law, changing stoichiometric coefficients multiplies ΔH\Delta H by the same factor, and reversing a reaction flips the sign of ΔH\Delta H.
3
Sum the enthalpy values for the target reaction
ΔHf∘=−788−286+1300=226 kJ mol−1\Delta H_f^\circ = -788 - 286 + 1300 = 226\text{ kJ mol}^{-1}
The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes for each intermediate step.

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Standard Enthalpy Changes and Hess's Law