If the internal energy change for the reaction
`N_(2(g))+3H_(2(g))hArr2NH_(3(g)`) is `-95(KJ)/(mol)` at `27^(@)` such that equilibrium is achieved at same temperature. Then calculate the `DeltaS_(surroundig)` ?

Options

A. `0`
B. `+(1000)/(3)Jk^(-1)mol^(-1)`
C. `-(1000)/(3)Jk^(-1)mol`
D. `-(1)/(3)Jk^(-1)mol^(-1)`

Explanation

To calculate the change in entropy of the surroundings (ΔS_surrounding) for the given reaction, we can follow these steps:

### Step 1: Write down the reaction and given data
The reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
Given:
- ΔU (internal energy change) = -95 kJ/mol
- Temperature (T) = 27°C = 300 K (since 27 + 273 = 300)

### Step 2: Calculate ΔH (enthalpy change)
Since the reaction is at constant temperature and pressure, we can relate ΔH to ΔU using the equation:
\[ \Delta H = \Delta U + \Delta n_g RT \]
Where:
- Δn_g = change in the number of moles of gas = (moles of products - moles of reactants)
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)

Calculate Δn_g:
- Moles of products (NH3) = 2
- Moles of reactants (N2 + 3H2) = 1 + 3 = 4
- Δn_g = 2 - 4 = -2

Now, substituting the values into the equation:
\[ \Delta H = -95 \text{ kJ/mol} + (-2) \times (0.008314 \text{ kJ/(mol·K)}) \times (300 \text{ K}) \]

Calculating:
\[ \Delta H = -95 \text{ kJ/mol} - 4.995 \text{ kJ/mol} \]
\[ \Delta H = -99.995 \text{ kJ/mol} \approx -100 \text{ kJ/mol} \]

### Step 3: Calculate ΔS_system
Using the Gibbs free energy equation at equilibrium:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, ΔG = 0, so:
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ T \Delta S = \Delta H \]
\[ \Delta S_{system} = \frac{\Delta H}{T} \]

Substituting the values:
\[ \Delta S_{system} = \frac{-100 \times 10^3 \text{ J/mol}}{300 \text{ K}} \]
\[ \Delta S_{system} = -333.33 \text{ J/(mol·K)} \]

### Step 4: Calculate ΔS_surrounding
The change in entropy of the surroundings is related to the system by:
\[ \Delta S_{surrounding} = -\Delta S_{system} \]

Substituting the value:
\[ \Delta S_{surrounding} = -(-333.33 \text{ J/(mol·K)}) \]
\[ \Delta S_{surrounding} = 333.33 \text{ J/(mol·K)} \]

### Final Answer
Thus, the change in entropy of the surroundings is:
\[ \Delta S_{surrounding} \approx 333.33 \text{ J/(mol·K)} \]

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