When charcoal burns in air signs of `DeltaH and DeltaS "are " 2C_((s)) + O_(2(g)) rarr 2CO_((g))`

Options

A. `underset(+)(DeltaH) underset(-)(DeltaS)`
B. `underset(-)(DeltaH) underset(+)(DeltaS)`
C. `underset(+)(DeltaH) underset(+)(DeltaS)`
D. `underset(-)(DeltaH) underset(-)(DeltaS)`

Answer

To determine the signs of ΔH (change in enthalpy) and ΔS (change in entropy) for the combustion of charcoal (represented by the reaction \(2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}\)), we can follow these steps:

### Step 1: Identify the nature of the reaction
The combustion of charcoal in air is an exothermic reaction, meaning it releases heat.

**Hint:** Exothermic reactions release heat, which typically results in a negative ΔH.

### Step 2: Determine the sign of ΔH
Since the reaction releases heat, the change in enthalpy (ΔH) is negative.

**Hint:** For exothermic reactions, ΔH is always negative.

### Step 3: Analyze the change in the number of moles of gas
In the given reaction, we have:
- Reactants: 2 moles of solid carbon (C) + 1 mole of oxygen gas (O₂) = 3 moles total
- Products: 2 moles of carbon monoxide gas (CO) = 2 moles total

**Hint:** Count the total number of moles of reactants and products to assess changes in the system.

### Step 4: Determine the change in entropy (ΔS)
The reaction shows a decrease in the total number of moles of gas (from 3 moles to 2 moles). A decrease in the number of moles generally indicates a decrease in randomness or disorder in the system.

**Hint:** A decrease in the number of moles of gas typically leads to a decrease in entropy.

### Step 5: Conclude the signs of ΔH and ΔS
From the analysis:
- ΔH is negative (exothermic reaction).
- ΔS is negative (decrease in randomness).

Thus, the signs of ΔH and ΔS for the combustion of charcoal are:
- ΔH < 0 (negative)
- ΔS < 0 (negative)

**Final Answer:** Both ΔH and ΔS are negative.