Consider the following reaction.
`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr6CO_(2)(g)+3H_(2)O(g)`
signs of `DeltaH, DeltaS` and `DeltaG` for the above reaction will be

Options

A. `+,-,+`
B. `-,+,-`
C. `-,+,-`
D. `+,+,-`

Explanation

To determine the signs of ΔH, ΔS, and ΔG for the given reaction:

**Reaction:**
\[ C_{6}H_{6}(l) + \frac{15}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(g) \]

### Step 1: Analyze ΔH (Enthalpy Change)
- This reaction is a combustion reaction of benzene (C6H6).
- Combustion reactions are typically exothermic, meaning they release heat.
- Therefore, the sign of ΔH for this reaction is **negative**.

### Step 2: Analyze ΔS (Entropy Change)
- We need to compare the number of moles of gaseous reactants and products.
- On the reactant side, we have:
- \( \frac{15}{2} O_{2} \) = 7.5 moles of gas.
- On the product side, we have:
- \( 6 CO_{2} + 3 H_{2}O \) = 9 moles of gas.
- Since the number of gaseous moles increases from 7.5 to 9, the entropy of the system increases.
- Therefore, the sign of ΔS for this reaction is **positive**.

### Step 3: Analyze ΔG (Gibbs Free Energy Change)
- The relationship between ΔG, ΔH, and ΔS is given by the equation:
\[ ΔG = ΔH - TΔS \]
- We have already established:
- ΔH is negative.
- ΔS is positive.
- Since ΔH is negative and ΔS is positive, the term \( -TΔS \) will also be negative (as T is always positive).
- Therefore, the overall ΔG will be negative (negative ΔH + negative \( -TΔS \)).
- Thus, the sign of ΔG for this reaction is **negative**.

### Summary of Signs:
- ΔH: Negative
- ΔS: Positive
- ΔG: Negative

### Final Answer:
- The signs of ΔH, ΔS, and ΔG for the reaction are:
- ΔH: Negative
- ΔS: Positive
- ΔG: Negative

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