- We manipulate the first equation by multiplying it by \(\frac{3}{2}\) to get \(3H_2(g) + \frac{3}{2}O_2(g) \rightarrow 3H_2O(g)\) with \(\Delta H = -725.4 \text{ kJ}\).
- We reverse the second equation and multiply by \(\frac{1}{2}\) to get \(O_3(g) \rightarrow \frac{3}{2}O_2(g)\) with \(\Delta H = -142.3 \text{ kJ}\).
- Summing these yields the target equation, and the total enthalpy change is \(-725.4 \text{ kJ} + (-142.3 \text{ kJ}) = -867.7 \text{ kJ}\).
- The final answer is \(\boxed{-867.7}\).
Explanation
- Analysis of the Thermochemical Equations
To find the overall enthalpy change for the target reaction \(3H_2(g) + O_3(g) \rightarrow 3H_2O(g)\), we apply Hess's Law using the provided thermochemical equations:
- \(2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \quad \Delta H_1 = -483.6 \text{ kJ}\)
- \(3O_2(g) \rightarrow 2O_3(g) \quad \Delta H_2 = 284.6 \text{ kJ}\)
Hess's Law states that if a reaction is the sum of several individual steps, the overall enthalpy change is the sum of the enthalpy changes of those steps. We must manipulate these equations so they sum to the target equation.
- Manipulating the Equations
To match the stoichiometric requirements of the target equation (\(3H_2(g) + O_3(g) \rightarrow 3H_2O(g)\)):
- Multiply the first equation by \(\frac{3}{2}\) to obtain 3 moles of \(H_2\) and 3 moles of \(H_2O\):
\(\frac{3}{2} \times [2H_2(g) + O_2(g) \rightarrow 2H_2O(g)] \implies 3H_2(g) + \frac{3}{2}O_2(g) \rightarrow 3H_2O(g)\)
The new enthalpy change is \(\Delta H_1' = \frac{3}{2} \times (-483.6 \text{ kJ}) = -725.4 \text{ kJ}\).
- Reverse the second equation and multiply by \(\frac{1}{2}\) to obtain 1 mole of \(O_3\) on the reactant side:
\(\frac{1}{2} \times [2O_3(g) \rightarrow 3O_2(g)] \implies O_3(g) \rightarrow \frac{3}{2}O_2(g)\)
The new enthalpy change is \(\Delta H_2' = -\frac{1}{2} \times (284.6 \text{ kJ}) = -142.3 \text{ kJ}\).
- Calculating the Overall Enthalpy Change
Now, add the manipulated equations:
\((3H_2(g) + \frac{3}{2}O_2(g)) + (O_3(g)) \rightarrow (3H_2O(g)) + (\frac{3}{2}O_2(g))\)
The \(\frac{3}{2}O_2(g)\) species cancels out on both sides, leaving:
\(3H_2(g) + O_3(g) \rightarrow 3H_2O(g)\)
The overall enthalpy change is the sum of the modified enthalpy changes:
\(\Delta H_{net} = \Delta H_1' + \Delta H_2' = -725.4 \text{ kJ} + (-142.3 \text{ kJ}) = -867.7 \text{ kJ}\).
The overall enthalpy of the reaction is \(-867.7 \text{ kJ}\).
Examples
Hess's Law is widely used in chemical engineering and industrial chemistry to calculate the heat involved in complex processes that cannot be measured directly. For example, if you want to calculate the heat released during the combustion of complex fuels in a furnace, you can use the known enthalpy values of simpler, intermediate reactions to predict the energy efficiency and safety requirements of the process without needing to perform the dangerous reaction itself. This ensures that large-scale industrial operations, like energy production or chemical synthesis, are designed with accurate thermal data.