Calculate `Delta_(r)S_("sys")^(@)` for the following reaction at 373 K:
`CO(g) + H_(2)O(g) to CO_(2)(g) + H_(2)(g)`
`Delta_(r)H^(@) = -4.1 xx 10^(4) J, Delta_(r)S^(@)("unv") = 56 J//K`

Options

A. `-54` J/K
B. `-166` J/K
C. `+54` J/K
D. `+166` J/K

Explanation

To calculate the change in entropy of the system (ΔS_sys) for the reaction at 373 K, we can follow these steps:

### Step 1: Identify Given Data
We have the following information:
- Reaction: \( CO(g) + H_2O(g) \rightarrow CO_2(g) + H_2(g) \)
- ΔH (enthalpy change) = -4.1 × 10^4 J
- ΔS (entropy change of the universe) = 56 J/K
- Temperature (T) = 373 K

### Step 2: Calculate ΔH of the Surroundings
The heat change of the surroundings (ΔH_surroundings) is equal to the negative of the heat change of the system (ΔH_sys):
\[ \Delta H_{\text{surroundings}} = -\Delta H_{\text{sys}} = -(-4.1 \times 10^4 \, \text{J}) = 4.1 \times 10^4 \, \text{J} \]

### Step 3: Calculate ΔS of the Surroundings
The change in entropy of the surroundings (ΔS_surroundings) can be calculated using the formula:
\[ \Delta S_{\text{surroundings}} = \frac{\Delta H_{\text{surroundings}}}{T} \]
Substituting the values:
\[ \Delta S_{\text{surroundings}} = \frac{4.1 \times 10^4 \, \text{J}}{373 \, \text{K}} \approx 110 \, \text{J/K} \]

### Step 4: Apply the Entropy Change of the Universe
The total change in entropy of the universe (ΔS_universe) is the sum of the change in entropy of the system and the surroundings:
\[ \Delta S_{\text{universe}} = \Delta S_{\text{sys}} + \Delta S_{\text{surroundings}} \]
Rearranging the equation to find ΔS_sys:
\[ \Delta S_{\text{sys}} = \Delta S_{\text{universe}} - \Delta S_{\text{surroundings}} \]
Substituting the known values:
\[ \Delta S_{\text{sys}} = 56 \, \text{J/K} - 110 \, \text{J/K} = -54 \, \text{J/K} \]

### Final Answer
The change in entropy of the system (ΔS_sys) for the reaction at 373 K is:
\[ \Delta S_{\text{sys}} = -54 \, \text{J/K} \]

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