From the following data at \(25^\circ C\), calculate \(\Delta_rH^\circ\) for \(H_2O(g)\longrightarrow 2H(g)+O(g)\).
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In Hess's law problems, reverse the reaction by changing the sign of \(\Delta H\), and multiply the reaction by multiplying \(\Delta H\) by the same factor.
Step 1: Identify the target reaction and data. We want $ \Delta_r H^\circ $ for: $ H_2O(g) \to 2H(g) + O(g) $. The data (from the image, standard values used): $ H_2(g) + \frac{1}{2}O_2(g) \to H_2O(g) $, $ \Delta H_1 = -242\text{ kJ mol}^{-1} $ (formation of water vapor); $ H_2(g) \to 2H(g) $, $ \Delta H_2 = 436\text{ kJ mol}^{-1} $ (H-H bond dissociation); $ O_2(g) \to 2O(g) $, $ \Delta H_3 = 496\text{ kJ mol}^{-1} $ (O=O bond dissociation). Step 2: Apply Hess's Law strategy. Hess's Law: the total enthalpy change is path-independent. We construct the target reaction by combining the given reactions algebraically. Step 3: Reverse the water formation reaction. Reversing reaction 1: $ H_2O(g) \to H_2(g) + \frac{1}{2}O_2(g) $, $ \Delta H = +242\text{ kJ mol}^{-1} $. This breaks water into H2 and O2. Step 4: Add the H2 dissociation reaction. Add reaction 2: $ H_2(g) \to 2H(g) $, $ \Delta H_2 = +436\text{ kJ mol}^{-1} $. Step 5: Add half of the O2 dissociation reaction. We need only 1 O atom, so take half of reaction 3: $ \frac{1}{2}O_2(g) \to O(g) $, $ \Delta H = +248\text{ kJ mol}^{-1} $. Now the sum is: $ H_2O(g) \to 2H(g) + O(g) $ ($ H_2 $ and $ \frac{1}{2}O_2 $ cancel from both sides). Step 6: Sum the enthalpy changes and state the answer. \[ \Delta_r H^\circ = 242 + 436 + 248 = 926\text{ kJ mol}^{-1} \] \[ \boxed{926\text{ kJ mol}^{-1}} \]
