Combustion of hydrogen in a fuel cell at 300 K is represented as `2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((g))`. If `Delta H` and `Delta G` are `-241.60 kJ mol^(-1)` and `-228.40 kJ mol^(-1)` of `H_(2)O`. The value of `DeltaS` for the above process is

Options

A. `4.4 JK^(-1)`
B. `-88 JK^(-1)`
C. `+88JK^(-1)`
D. `-44JK^(-1)`

Answer

To find the value of the entropy change (ΔS) for the combustion of hydrogen in a fuel cell, we can use the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS). The reaction given is:

\[ 2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)} \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- ΔH = -241.60 kJ/mol (for the formation of 2 moles of water)
- ΔG = -228.40 kJ/mol (for the formation of 2 moles of water)
- Temperature (T) = 300 K

2. **Use the Gibbs Free Energy Equation:**
The relationship between ΔG, ΔH, and ΔS is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
Rearranging this equation to find ΔS gives:
\[ T \Delta S = \Delta H - \Delta G \]

3. **Substitute the Values:**
Substitute the values of ΔH and ΔG into the equation:
\[ T \Delta S = (-241.60 \text{ kJ/mol}) - (-228.40 \text{ kJ/mol}) \]
\[ T \Delta S = -241.60 + 228.40 = -13.20 \text{ kJ/mol} \]

4. **Convert kJ to Joules:**
Since 1 kJ = 1000 J, we convert -13.20 kJ to Joules:
\[ T \Delta S = -13.20 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -13200 \text{ J/mol} \]

5. **Calculate ΔS:**
Now, divide by the temperature (T = 300 K) to find ΔS:
\[ \Delta S = \frac{-13200 \text{ J/mol}}{300 \text{ K}} = -44 \text{ J/(mol·K)} \]

6. **Adjust for the Reaction Stoichiometry:**
The ΔS calculated is for the formation of 1 mole of water. Since the reaction produces 2 moles of water, we multiply ΔS by 2:
\[ \Delta S = -44 \text{ J/(mol·K)} \times 2 = -88 \text{ J/(mol·K)} \]

### Final Answer:
The value of ΔS for the combustion of hydrogen in the fuel cell is:
\[ \Delta S = -88 \text{ J/(mol·K)} \]