Question:
From the following data at \(25^\circ C\), calculate \(\Delta_rH^\circ\) for \(H_2O(g)\longrightarrow 2H(g)+O(g)\).
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.
Updated On: Jun 25, 2026
- \(1174\ kJ\ mol^{-1}\)
- \(742\ kJ\ mol^{-1}\)
- \(926\ kJ\ mol^{-1}\)
- \(690\ kJ\ mol^{-1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Write the required reaction.
We have to calculate enthalpy change for \[ H_2O(g)\longrightarrow 2H(g)+O(g) \]
Step 2: Reverse the formation reaction of water.
Given, \[ H_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(g) \] \[ \Delta H=-242\ kJ\ mol^{-1} \] On reversing, \[ H_2O(g)\longrightarrow H_2(g)+\frac{1}{2}O_2(g) \] \[ \Delta H=+242\ kJ\ mol^{-1} \]
Step 3: Dissociate hydrogen molecule into hydrogen atoms.
Given, \[ H_2(g)\longrightarrow 2H(g) \] \[ \Delta H=436\ kJ\ mol^{-1} \]
Step 4: Dissociate oxygen molecule to get one oxygen atom.
Given, \[ O_2(g)\longrightarrow 2O(g) \] \[ \Delta H=496\ kJ\ mol^{-1} \] Therefore, \[ \frac{1}{2}O_2(g)\longrightarrow O(g) \] \[ \Delta H=\frac{496}{2}=248\ kJ\ mol^{-1} \]
Step 5: Add all enthalpy changes.
\[ \Delta_rH^\circ=242+436+248 \] \[ \Delta_rH^\circ=926\ kJ\ mol^{-1} \]
Step 6: Final conclusion.
Hence, \[ \boxed{926\ kJ\ mol^{-1}} \]
We have to calculate enthalpy change for \[ H_2O(g)\longrightarrow 2H(g)+O(g) \]
Step 2: Reverse the formation reaction of water.
Given, \[ H_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(g) \] \[ \Delta H=-242\ kJ\ mol^{-1} \] On reversing, \[ H_2O(g)\longrightarrow H_2(g)+\frac{1}{2}O_2(g) \] \[ \Delta H=+242\ kJ\ mol^{-1} \]
Step 3: Dissociate hydrogen molecule into hydrogen atoms.
Given, \[ H_2(g)\longrightarrow 2H(g) \] \[ \Delta H=436\ kJ\ mol^{-1} \]
Step 4: Dissociate oxygen molecule to get one oxygen atom.
Given, \[ O_2(g)\longrightarrow 2O(g) \] \[ \Delta H=496\ kJ\ mol^{-1} \] Therefore, \[ \frac{1}{2}O_2(g)\longrightarrow O(g) \] \[ \Delta H=\frac{496}{2}=248\ kJ\ mol^{-1} \]
Step 5: Add all enthalpy changes.
\[ \Delta_rH^\circ=242+436+248 \] \[ \Delta_rH^\circ=926\ kJ\ mol^{-1} \]
Step 6: Final conclusion.
Hence, \[ \boxed{926\ kJ\ mol^{-1}} \]
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