Following reaction occurring in an automobile `2C_(8)H_(18)(g)+25O_(2)(g) to 16CO_(2)(g)+18H_(2)O(g)`. The sign of `triangleH, triangleS and triangleG` would be-

Options

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

Answer

To determine the signs of ΔH, ΔS, and ΔG for the given combustion reaction of octane, we will analyze each term step by step.

### Given Reaction:
\[ 2C_{8}H_{18}(g) + 25O_{2}(g) \rightarrow 16CO_{2}(g) + 18H_{2}O(g) \]

### Step 1: Determine the sign of ΔH (Change in Enthalpy)
- The reaction is a combustion reaction, which is an exothermic process.
- In exothermic reactions, heat is released to the surroundings.
- Therefore, the change in enthalpy (ΔH) is negative.

**Conclusion for ΔH:**
\[ \Delta H < 0 \]

### Step 2: Determine the sign of ΔS (Change in Entropy)
- Entropy (S) is a measure of disorder in a system.
- We need to compare the number of gaseous reactants and products.
- **Reactants:**
- 2 moles of \(C_{8}H_{18}(g)\) + 25 moles of \(O_{2}(g)\) = 27 moles of gaseous reactants.
- **Products:**
- 16 moles of \(CO_{2}(g)\) + 18 moles of \(H_{2}O(g)\) = 34 moles of gaseous products.
- Since the number of gaseous products (34) is greater than the number of gaseous reactants (27), the disorder of the system increases.
- Therefore, the change in entropy (ΔS) is positive.

**Conclusion for ΔS:**
\[ \Delta S > 0 \]

### Step 3: Determine the sign of ΔG (Change in Gibbs Free Energy)
- The Gibbs free energy change (ΔG) can be calculated using the equation:
\[ \Delta G = \Delta H - T\Delta S \]
- We know:
- ΔH is negative (ΔH < 0)
- ΔS is positive (ΔS > 0)
- Since T (temperature) is always positive, the term \(T\Delta S\) will also be positive.
- Therefore, when we subtract a positive value from a negative value, ΔG will also be negative.

**Conclusion for ΔG:**
\[ \Delta G < 0 \]

### Final Results:
- ΔH: Negative
- ΔS: Positive
- ΔG: Negative

### Summary:
- **ΔH**: Negative (exothermic reaction)
- **ΔS**: Positive (increase in disorder)
- **ΔG**: Negative (reaction is spontaneous)