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 given reaction at 373 K, we can follow these steps:

### Step 1: Write down the given data
We have the following information:
- Reaction: \( \text{CO(g)} + \text{H}_2\text{O(g)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \)
- \( \Delta_rH^\circ = -4.1 \times 10^4 \, \text{J} \)
- \( \Delta_rS^\circ(\text{univ}) = 56 \, \text{J/K} \)
- Temperature (T) = 373 K

### Step 2: Use the relationship between ΔS_univ, ΔS_sys, and ΔS_surr
The total change in entropy of the universe (ΔS_univ) is given by the sum of the change in entropy of the system (ΔS_sys) and the change in entropy of the surroundings (ΔS_surr):
\[ \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \]

### Step 3: Calculate ΔS_surr
To find ΔS_surr, we can use the formula:
\[ \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T} \]
Since the reaction releases heat (ΔH is negative), we take the absolute value of ΔH:
\[ \Delta S_{\text{surr}} = -\frac{-4.1 \times 10^4 \, \text{J}}{373 \, \text{K}} = \frac{4.1 \times 10^4 \, \text{J}}{373 \, \text{K}} \]

### Step 4: Perform the calculation for ΔS_surr
Calculating ΔS_surr:
\[ \Delta S_{\text{surr}} \approx \frac{41000}{373} \approx 110.0 \, \text{J/K} \]

### Step 5: Substitute ΔS_surr into the equation for ΔS_univ
Now, we can substitute ΔS_surr back into the equation for ΔS_univ:
\[ 56 \, \text{J/K} = \Delta S_{\text{sys}} + 110.0 \, \text{J/K} \]

### Step 6: Solve for ΔS_sys
Rearranging the equation to find ΔS_sys:
\[ \Delta S_{\text{sys}} = 56 \, \text{J/K} - 110.0 \, \text{J/K} = -54 \, \text{J/K} \]

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

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