What will be the value of `triangle_(r)S_(sys)^(@)` for the following reaction at 373 K: `CO(g)+H_(2)O(g) to CO_(2)(g)+H_(2)(g) triangle_(r) H^(@)=-4.5 xx 10^(4) J, triangle_(r)S_("univ")^(@) =57.5J/K`

Options

A. `+178J/K`
B. `-166J/K`
C. `-178J/K`
D. `+166 J/K`

Answer

To find the value of the standard change in entropy of the system (ΔS_sys^(@)) for the reaction at 373 K, we can use the following steps:

### Step 1: Write down the given values.
- ΔH^(@) = -4.5 × 10^4 J (standard change in enthalpy)
- ΔS_univ^(@) = 57.5 J/K (standard change in entropy of the universe)
- T = 373 K (temperature)

### Step 2: Use the relationship between the entropy of the universe, the system, and the surroundings.
According to the second law of thermodynamics:
\[ ΔS_{univ}^(@) = ΔS_{sys}^(@) + ΔS_{surr}^(@) \]
Where:
- ΔS_{sys}^(@) = standard change in entropy of the system
- ΔS_{surr}^(@) = standard change in entropy of the surroundings

### Step 3: Calculate ΔS_{surr}^(@).
The change in entropy of the surroundings can be calculated using the formula:
\[ ΔS_{surr}^(@) = -\frac{ΔH^(@)}{T} \]
Substituting the known values:
\[ ΔS_{surr}^(@) = -\frac{-4.5 \times 10^4 \text{ J}}{373 \text{ K}} \]
\[ ΔS_{surr}^(@) = \frac{4.5 \times 10^4}{373} \]
Calculating this gives:
\[ ΔS_{surr}^(@) ≈ 120.64 \text{ J/K} \]

### Step 4: Substitute ΔS_{surr}^(@) into the equation for ΔS_{univ}^(@).
Now we can substitute ΔS_{surr}^(@) back into the equation for ΔS_{univ}^(@):
\[ 57.5 \text{ J/K} = ΔS_{sys}^(@) + 120.64 \text{ J/K} \]

### Step 5: Solve for ΔS_{sys}^(@).
Rearranging the equation to isolate ΔS_{sys}^(@):
\[ ΔS_{sys}^(@) = 57.5 \text{ J/K} - 120.64 \text{ J/K} \]
Calculating this gives:
\[ ΔS_{sys}^(@) = -63.14 \text{ J/K} \]

### Final Answer:
The value of \( ΔS_{sys}^(@) \) for the reaction at 373 K is approximately **-63.14 J/K**.

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