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. `Delta H = T Delta S`
B. `Delta H gt T Delta S`
C. `Delta H lt T Delta S`
D. `Delta H gt 0 and Tdelta S lt 0`

Answer

To determine the conditions under which the reaction

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

proceeds spontaneously at \( 27^{\circ}C \), we will analyze the Gibbs free energy change (\( \Delta G \)) for the reaction.

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation**:
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- \( \Delta G \) = change in Gibbs free energy
- \( \Delta H \) = change in enthalpy
- \( T \) = temperature in Kelvin
- \( \Delta S \) = change in entropy

2. **Identify the Nature of the Reaction**:
The given reaction is a combustion reaction, which typically releases heat. Therefore, we can conclude that the reaction is exothermic, meaning:
\[ \Delta H < 0 \]

3. **Calculate the Change in Entropy (\( \Delta S \))**:
To determine the sign of \( \Delta S \), we need to compare the number of moles of gaseous reactants and products.
- Reactants: \( \frac{15}{2} O_2 = 7.5 \) moles of \( O_2 \) (gaseous)
- Products: \( 6 CO_2 + 3 H_2O \) (only \( CO_2 \) is gaseous, which is 6 moles)

Since there are more moles of gaseous reactants (7.5 moles) than gaseous products (6 moles), the entropy change will be:
\[ \Delta S < 0 \]

4. **Determine the Conditions for Spontaneity**:
For the reaction to be spontaneous, \( \Delta G \) must be negative:
\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]
Rearranging gives:
\[ \Delta H < T \Delta S \]

5. **Substituting the Signs**:
Since \( \Delta H < 0 \) and \( \Delta S < 0 \), we can substitute these into the inequality:
\[ \text{(Negative)} < T \times \text{(Negative)} \]
This implies that:
\[ \Delta H > T \Delta S \]
Thus, for the reaction to be spontaneous, the absolute value of \( \Delta H \) must be greater than the product of temperature and \( \Delta S \).

### Conclusion:
The condition for the reaction to proceed spontaneously at \( 27^{\circ}C \) is that \( \Delta H \) must be greater than \( T \Delta S \).