Select correct statement(s) for the reaction `H_(2)O(g)+CO(g)rarrH_(2)(g)+CO_(2)(g)` substance
`CO(g)" "CO_(2)(g)" "H_(2)O(g)" "H_(2)(g)`
`Delta_(f)H_(400)^(@)("kcal mol"^(-1))" -25 -95 -55 0"`
`S_(400)^(@)("cal mol"^(-1)"K"^(-1)) " 45 50 40 30"`

Options

A. Reaction is enthalpy driven
B. Reaction is entropy driven
C. Reaction is spontaneous at 400 K
D. Reaction is non-spontaneous at 400 K

Explanation

To solve the problem, we need to calculate the change in enthalpy (ΔH) and the change in entropy (ΔS) for the given reaction:

**Reaction:**
\[ H_2O(g) + CO(g) \rightarrow H_2(g) + CO_2(g) \]

### Step 1: Calculate ΔH for the reaction

The change in enthalpy (ΔH) for the reaction can be calculated using the standard enthalpy of formation (ΔH_f) values provided:

\[ \Delta H_{reaction} = \Delta H_f(H_2) + \Delta H_f(CO_2) - \Delta H_f(H_2O) - \Delta H_f(CO) \]

Substituting the values from the question:

- ΔH_f(H2) = 0 kcal/mol (by definition)
- ΔH_f(CO2) = -95 kcal/mol
- ΔH_f(H2O) = -55 kcal/mol
- ΔH_f(CO) = -25 kcal/mol

Now substituting these values into the equation:

\[ \Delta H_{reaction} = 0 + (-95) - (-55) - (-25) \]

Calculating further:

\[ \Delta H_{reaction} = 0 - 95 + 55 + 25 \]
\[ \Delta H_{reaction} = -95 + 80 = -15 \text{ kcal/mol} \]

### Step 2: Calculate ΔS for the reaction

The change in entropy (ΔS) for the reaction can be calculated using the standard entropy (S) values provided:

\[ \Delta S_{reaction} = S(H_2) + S(CO_2) - S(H_2O) - S(CO) \]

Substituting the values from the question:

- S(H2) = 30 cal/mol·K
- S(CO2) = 50 cal/mol·K
- S(H2O) = 40 cal/mol·K
- S(CO) = 45 cal/mol·K

Now substituting these values into the equation:

\[ \Delta S_{reaction} = 30 + 50 - 40 - 45 \]

Calculating further:

\[ \Delta S_{reaction} = 80 - 85 = -5 \text{ cal/mol·K} \]

### Step 3: Determine the spontaneity of the reaction

To determine if the reaction is spontaneous, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- ΔH = -15 kcal/mol (convert to cal: -15,000 cal/mol)
- T = 400 K
- ΔS = -5 cal/mol·K

Substituting the values into the Gibbs free energy equation:

\[ \Delta G = -15000 - (400)(-5) \]
\[ \Delta G = -15000 + 2000 \]
\[ \Delta G = -13000 \text{ cal/mol} \]

Since ΔG is negative, the reaction is spontaneous at 400 K.

### Conclusion

The correct statements regarding the reaction are:
1. The reaction is exothermic (ΔH < 0).
2. The reaction has a negative change in entropy (ΔS < 0).
3. The reaction is spontaneous at 400 K (ΔG < 0).