At `27^(@)C` for reaction,
`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l)`
proceeds spontaneously because the magnitude of

Options

A. `DeltaH = T Delta S`
B. `DeltaH gt T Delta S`
C. `DeltaH lt T Delta S`
D. `DeltaH gt 0` and `T Delta S lt 0`

Explanation

To solve the problem, we need to analyze the given reaction and the thermodynamic principles involved. The reaction is:

\[ C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l) \]

### Step 1: Understanding the Reaction
This reaction represents the combustion of benzene (C6H6) in the presence of oxygen (O2), producing carbon dioxide (CO2) and water (H2O).

### Step 2: Determine the Signs of ΔH and ΔS
- **ΔH (Enthalpy Change)**: For combustion reactions, ΔH is typically negative because energy is released when the bonds in the reactants are broken and new bonds are formed in the products.
- **ΔS (Entropy Change)**: To determine the sign of ΔS, we need to compare the number of gaseous moles on the reactant side and the product side.

### Step 3: Count the Moles of Gas
- **Reactants**:
- 1 mole of C6H6 (liquid, does not contribute to gas moles)
- \( \frac{15}{2} \) moles of O2 (gas)
- Total gas moles = \( \frac{15}{2} = 7.5 \) moles

- **Products**:
- 6 moles of CO2 (gas)
- 3 moles of H2O (liquid, does not contribute to gas moles)
- Total gas moles = 6 moles of CO2

### Step 4: Calculate ΔS
- The total moles of gas in the reactants = 7.5 moles
- The total moles of gas in the products = 6 moles

Since there are fewer gaseous moles in the products than in the reactants, ΔS is negative.

### Step 5: Apply the Gibbs Free Energy Equation
The Gibbs free energy change (ΔG) is given by:

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

For the reaction to be spontaneous, ΔG must be negative:

\[ \Delta G < 0 \]

This implies:

\[ \Delta H - T \Delta S < 0 \]

Rearranging gives:

\[ \Delta H < T \Delta S \]

### Step 6: Analyze the Conditions for Spontaneity
Since ΔH is negative and ΔS is negative, for the reaction to be spontaneous, we need:

\[ \Delta H > T \Delta S \]

This means that the magnitude of ΔH must be greater than the magnitude of TΔS (considering that TΔS is a negative value).

### Conclusion
The correct option is:

**ΔH > TΔS**