Question

At \(25^{\circ} C\), the enthalpy of the following processes are given: 
\(H _2( g )+ O _2( g ) \rightarrow 2 OH ( g ) \Delta H ^{\circ} =78 \,kJ \,mol ^{-1}\)
\(H _2( g )+1 / 2 O _2( g ) \rightarrow H _2 O ( g ) \Delta H ^{\circ} =-242 \,kJ \,mol ^{-1}\)
\(H _2( g ) \rightarrow 2 H ( g ) \Delta H ^{\circ} =436\, kJ \,mol ^{-1}\)
\(1 / 2 O _2( g ) \rightarrow O ( g ) \Delta H ^{\circ} =249\, kJ\, mol ^{-1}\) 
What would be the value of \(X\) for the following reaction?______ (Nearest integer) 
\(H _2 O ( g ) \rightarrow H ( g )+ OH ( g ) \Delta H ^{\circ}= X\, kJ\, mol -1\)

Hint:

Hess’s Law allows us to calculate the enthalpy change of a reaction by summing the enthalpy changes of individual steps that lead to the overall reaction

Answer 1

Correct Answer: 499

Calculation of Enthalpy Change for H₂O(g) → H(g) + OH(g): 

Using Hess’s Law, we can derive the enthalpy change for this reaction by breaking it into steps:

1. Step 1: Break H₂O(g) into its constituent atoms:
   Reaction: H₂O(g) → 2H(g) + O(g)
   From the given data:
   ∆H° = ∆H°_{H2 → 2H} + ∆H°_{1/2 O2 → O}
   Substituting the values:
   ∆H° = 436 + 249 = 685 kJ mol⁻¹.
2. Step 2: Combine 1 hydrogen atom (H(g)) and 1 oxygen atom (O(g)) to form an OH radical:
   Reaction: H(g) + O(g) → OH(g)
   Using the reverse of the reaction H₂(g) + O₂(g) → 2OH(g), where ∆H° = 78 kJ mol⁻¹:
   Divide by 2 to get the enthalpy change for forming 1 mole of OH:
   ∆H° = 78 / 2 = 39 kJ mol⁻¹.
3. Step 3: Combine the results:
   Total enthalpy change:
   X = 685 − 39 = 499 kJ mol⁻¹.

Conclusion: The enthalpy change for the reaction H₂O(g) → H(g) + OH(g) is 499 kJ mol⁻¹.

Answer 2

correct answer is 499.