For a cell reaction `2H_2(g)+O_2(g)to2H_2O(l)/_\_rS_(198)^(@)=-0.32KJ//k` . What is the value of `/_\_fH_(298)^(@)(H_2O,l)` ?
Given: `O_2(g)+4H^+(aq)+4e^-)to2H_2O(l), E^(@)=1.23V`

Options

A. `-285.07kJ//mol`
B. `-570.14kJ//mol`
C. `285.kJ//mol`
D. None of these

Answer

To find the value of the standard enthalpy of formation \((\Delta_f H_{298}^\circ)\) for water \((H_2O, l)\), we can follow these steps:

### Step 1: Write down the relevant equations
We know that:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Rearranging this gives us:
\[ \Delta H^\circ = \Delta G^\circ + T \Delta S^\circ \]

### Step 2: Calculate \(\Delta G^\circ\)
We can calculate \(\Delta G^\circ\) using the formula:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \(n = 4\) (the number of moles of electrons transferred)
- \(F = 96500 \, C/mol\) (Faraday's constant)
- \(E^\circ = 1.23 \, V\) (the standard cell potential)

Substituting the values:
\[ \Delta G^\circ = -4 \times 96500 \times 1.23 \]
Calculating this gives:
\[ \Delta G^\circ = -474.78 \, kJ/mol \]

### Step 3: Substitute values into the enthalpy equation
Now we substitute \(\Delta G^\circ\) and \(\Delta S^\circ\) into the enthalpy equation. We have:
- \(\Delta S^\circ = -0.32 \, kJ/K\) (which is \(-320 \, J/K\))
- \(T = 298 \, K\)

Now, substituting these values:
\[ \Delta H^\circ = -474.78 \, kJ/mol + 298 \times (-0.32 \, kJ/K) \]
Calculating \(298 \times -0.32\):
\[ 298 \times -0.32 = -95.36 \, kJ \]
Now substituting this back:
\[ \Delta H^\circ = -474.78 \, kJ/mol - 95.36 \, kJ \]
\[ \Delta H^\circ = -570.14 \, kJ/mol \]

### Step 4: Adjust for the formation of one mole of water
Since the reaction given is for the formation of 2 moles of water, we need to divide the enthalpy by 2 to find the standard enthalpy of formation for one mole of water:
\[ \Delta_f H_{298}^\circ (H_2O, l) = \frac{-570.14 \, kJ/mol}{2} = -285.07 \, kJ/mol \]

### Final Answer
Thus, the value of \(\Delta_f H_{298}^\circ (H_2O, l)\) is:
\[ \Delta_f H_{298}^\circ (H_2O, l) = -285.07 \, kJ/mol \]

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